Question 15 Marks
$\frac{\text{c}}{\text{a}-\text{b}}=\frac{\tan\big(\frac{\text{A}}{2}\big)+\tan\big(\frac{\text{B}}{2}\big)}{\tan\big(\frac{\text{A}}{2}\big)-\tan\big(\frac{\text{B}}{2}\big)}$
Answer$\frac{\text{c}}{\text{a}-\text{b}}=\frac{\tan\big(\frac{\text{A}}{2}\big)+\tan\big(\frac{\text{B}}{2}\big)}{\tan\big(\frac{\text{A}}{2}\big)-\tan\big(\frac{\text{B}}{2}\big)}$$\text{LHS}=\frac{\text{c}}{\text{a}-\text{b}}$
$=\frac{\text{k}\sin\text{C}}{\text{k}\sin\text{A}-\text{k}\sin\text{B}}$
$=\frac{\text{}\sin\text{C}}{\text{}\sin\text{A}-\text{}\sin\text{B}}$
$=\frac{2\sin\frac{\text{C}}{2}\cos\frac{\text{C}}{2}}{\sin\text{A}-\sin\text{B}}$
$=\frac{2\sin\frac{\text{C}}{2}\cos\frac{\text{C}}{2}}{2\cos\big(\frac{\text{A +B}}{2}\big)\sin\big(\frac{\text{A}-\text{B}}{2}\big)}$
$=\frac{\sin\frac{\text{C}}{2}\cos\frac{(\pi-(\text{A + B})}{2}}{\cos\big(\frac{\pi-\text{C}}{2}\big)\sin\big(\frac{\text{A}-\text{B}}{2}\big)}$
$=\frac{\sin\frac{\text{C}}{2}\sin\frac{(\text{A + B})}{2}}{\sin\frac{\text{C}}{2}\sin\big(\frac{\text{A}-\text{B}}{2}\big)}$
$=\frac{\sin\frac{\text{(A + B)}}{2}}{\sin\frac{\text{(A - B)}}{2}}$
$=\frac{\sin\big(\frac{\text{A}}{2}\big).\cos\big(\frac{\text{B}}{2}\big)+\sin\big(\frac{\text{B}}{2}\big).\cos\big(\frac{\text{A}}{2}\big)}{\sin\big(\frac{\text{A}}{2}\big).\cos\big(\frac{\text{B}}{2}\big)-\sin\big(\frac{\text{B}}{2}\big).\cos\big(\frac{\text{A}}{2}\big)}$
$=\frac{\tan\big(\frac{\text{A}}{2}\big)+\tan\big(\frac{\text{B}}{2}\big)}{\tan\big(\frac{\text{A}}{2}\big).-\tan\big(\frac{\text{B}}{2}\big).}=\text{RHS}$ $\Big[$Dividing both Numerator and Denominator by $\cos\big(\frac{\text{A}}{2}\big).\cos\big(\frac{\text{B}}{2}\big)\Big]$
View full question & answer→Question 25 Marks
In any $\triangle\text{ABC},\frac{\text{b + c}}{12}=\frac{\text{c + a}}{13}=\frac{\text{a + b}}{15},$ then prove that $\frac{\cos\text{A}}{2}=\frac{\cos\text{B}}{7}=\frac{\cos\text{C}}{11}.$
AnswerLet $\frac{\text{b + c}}{12}=\frac{\text{c + a}}{13}=\frac{\text{a + b}}{15}=\lambda\text{(say)}$
$\text{b + c}=12\lambda,\text{c + a}=13\lambda,\text{a + b}=15\lambda$
$(\text{b + c + c + a + a + b})=12\lambda+13\lambda+15\lambda$
$2(\text{a + b + c})=40\lambda$
$\text{a + b + c}=20\lambda$
$\text{b + c}=12\lambda$ and $\text{a + b + c}=20\lambda\Rightarrow\text{a}=8\lambda$
$\text{c + a}=13\lambda$ and $\text{a + b + c}=20\lambda\Rightarrow\text{b}=7\lambda$
$\text{a + b}=15\lambda$ and $\text{a + b + c}=20\lambda\Rightarrow\text{c}=5\lambda$
$\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}=\frac{49\lambda^2+25\lambda^2-64\lambda^2}{70\lambda^2}=\frac{1}{7}$
$\cos\text{B}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}=\frac{64\lambda^2+25\lambda^2-49\lambda^2}{80\lambda^2}=\frac{1}{2}$
$\cos\text{C}=\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}=\frac{64\lambda^2+49\lambda^2-25\lambda^2}{112\lambda^2}=\frac{11}{14}$
$\cos\text{A}:\cos\text{B}:\cos\text{C}=\frac{1}{7}:\frac{1}{2}:\frac{11}{14}=2:7:11$
View full question & answer→Question 35 Marks
$\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)=\frac{\text{b}-\text{c}}{\text{a}}\cos\frac{\text{A}}{2}$
AnswerThen,
Consider the RHS of the equation $\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)=\frac{\text{b}-\text{c}}{\text{a}}\cos\frac{\text{A}}{2}$
$\text{RHS}=\frac{\text{b}-\text{c}}{\text{a}}\cos\frac{\text{A}}{2}$
$=\frac{\text{k}(\sin\text{B}-\sin\text{C})}{\text{k}\sin\text{A}}\cos\Big(\frac{\pi-(\text{B + C})}{2}\Big)$ $(\because\text{A + B + C}=\pi)$
$=\frac{2\sin\big(\frac{\text{B}-\text{C}}{2}\big)\cos\big(\frac{\text{B + C}}{2}\big)}{\sin\text{A}}$
$=\frac{\sin\big(\frac{\text{B}-\text{C}}{2}\big)2\cos\big(\frac{\text{B + C}}{2}\big)}{\sin\text{A}}\sin\Big(\frac{\text{B + C}}{2}\Big)$
$=\frac{\sin\big(\frac{\text{B}-\text{C}}{2}\big)\sin(\text{B + C})}{\sin\text{A}}$
$=\frac{\sin\big(\frac{\text{B} - \text{C}}{2}\big)\sin(\pi-\text{A})}{\sin\text{A}}$
$=\frac{\sin\text{A}\sin\big(\frac{\text{B}-\text{C}}{2}\big)}{\sin\text{A}}$
$=\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)=\text{LHS}$
Hence proved.
View full question & answer→Question 45 Marks
$\text{a}\cos\text{A + b}\cos\text{B + c}\cos\text{C = 2b}\sin\text{A}\sin\text{C}$
AnswerBy sine rule, we know that
$\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}=\text{k }\text{(say)}$
$\Rightarrow\text{a = k}\sin\text{A, b = k}\sin\text{B, c = k}\sin\text{C}$
Now,
$\text{LHS}=\text{a}\cos\text{A + b}\cos\text{B + c}\cos\text{C}$
$=\text{k}\sin\text{A}\cos\text{A + k}\sin\text{B}\cos\text{B + k}\sin\text{C}\cos\text{C}$
$=\frac{\text{k}}{2}(2\sin\text{A}\cos\text{A}+2\sin\text{B}\cos\text{B}+2\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}(\sin2\text{A}+\sin2\text{B}+2\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}\Big(2\sin\frac{2\text{A}+2\text{B}}{2}\cos\frac{2\text{A}-2\text{B}}{2}+2\sin\text{C}\cos\text{C}\Big)$
$=\frac{\text{k}}{2}(2\sin(\text{A + B})\cos(\text{A}-\text{B})+2\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}(2\sin(\pi-\text{C})\cos(\text{A}-\text{B})+2\sin\text{C}\cos\text{C})$ $(\because\text{A + B + C}=\pi)$
$=\frac{\text{k}}{2}(2\sin\text{C}\cos(\text{A}-\text{B})+2\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}\times2\sin\text{C}(\cos(\text{A}-\text{B})+\cos\text{C})$
$=\text{k}\sin\text{C}\Big(2\cos\Big(\frac{\text{A}-\text{B + C}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}-\text{C}}{2}\Big)\Big)$
$=\text{k}\sin\text{C}\Big(2\cos\Big(\frac{\pi-\text{B}-\text{B}}{2}\Big)\cos\Big(\frac{\text{B}+\text{C}-\text{A}}{2}\Big)\Big)$ $(\because\text{A + B + C}=\pi)$
$=\text{k}\sin\text{C}\Big(2\cos\Big(\frac{\pi-2\text{B}}{2}\Big)\cos\Big(\frac{\pi-2\text{A}}{2}\Big)\Big)$ $(\because\text{A + B + C}=\pi)$
$=\text{k}\sin\text{C}\Big(2\cos\Big(\frac{\pi}{2}-\text{B}\Big)\cos\Big(\frac{\pi}{2}-\text{A}\Big)\Big)$
$=2\text{k}\sin\text{C}(\sin\text{B}\sin\text{A})$
$=2(\text{k}\sin\text{B})\sin\text{A}\sin\text{C}$
$=2\text{b}\sin\text{A}\sin\text{C}$
$=\text{RHS}$
$\therefore\text{LHS = RHS}$
Hence, $\text{a}\cos\text{A + b}\cos\text{B + c}\cos\text{C = 2 b}\sin\text{A}\sin\text{C}.$
View full question & answer→Question 55 Marks
$\text{a}^2=(\text{b + c})^2-4\text{bc}\cos^2\frac{\text{A}}{2}$
Answer$\text{RHS}=(\text{b + c})^2-4\text{bc}\cos^2\frac{\text{A}}{2}$
$=\text{b}^2+\text{c}^2+2\text{bc}-4\text{bc}\Big(\frac{1+\cos\text{A}}{2}\Big)$
$=\text{b}^2+\text{c}^2+2\text{bc}-2\text{bc}(1+\cos\text{A})$
$=\text{b}^2+\text{c}^2+2\text{bc}(1-1-\cos\text{A})$
$=\text{b}^2+\text{c}^2-2\text{bc}\cos\text{A}$
$=\text{b}^2+\text{c}^2-2\text{bc}\Big(\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}\Big)$ $\Big(\because\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}\Big)$
$=\text{b}^2+\text{c}^2-\text{b}^2-\text{c}^2+\text{a}^2$
$=\text{a}^2=\text{LHS}$
Hence proved.
View full question & answer→Question 65 Marks
The upper part of a tree broken by the wind makes an angle of 30° with the ground and the distance from the root to the point where the top of the tree touches the ground is 15m. Using sine rule, find the height of the tree.
AnswerSuppose BD be the tree and the upper part of the tree is broken over by the wind at point A.
The total height of the tree is x + y.
In $\triangle\text{ABC},$
$\angle\text{C}=30^{\circ}$ and $\angle\text{B}=90^{\circ}.$
$\therefore\angle\text{A}=60^{\circ}.$
So, on using sine rule, we get:
$\frac{\text{AB}}{\sin30^{\circ}}=\frac{\text{BC}}{\sin60^{\circ}}=\frac{\text{AC}}{\sin90^{\circ}}$
$\Rightarrow\frac{\text{x}}{\sin30^{\circ}}=\frac{15}{\sin60^{\circ}}=\frac{\text{y}}{\sin90^{\circ}}$
So, $\frac{\text{x}}{\sin30^{\circ}}=\frac{15}{\sin60^{\circ}}$
$\Rightarrow\frac{\text{x}}{\frac{1}{2}}=\frac{15}{\frac{\sqrt{3}}{2}}$
$\Rightarrow\text{x}=\frac{15}{\sqrt{3}}=5\sqrt{3}$
Also,
$\frac{15}{\sin60^{\circ}}=\frac{\text{y}}{\sin90^{\circ}}$
$\Rightarrow\frac{15}{\frac{\sqrt{3}}2{}}=\text{y}$
$\Rightarrow\text{y}=\frac{30}{\sqrt{3}}=10\sqrt{3}$
So, the height of the tree is $\text{x + y}=5\sqrt{3}+10\sqrt{3}\text{m}$
$=15\sqrt{3}\text{m}$ View full question & answer→Question 75 Marks
In any $\triangle A B C$, if $a^2, b^2, c^2$ are in A.P., prove that $\cot A, \cot B$ and $\cot C$ are also in A.P.
Answer$a^2, b^2, c^2$ are in A.P.$\Rightarrow-2\text{a}^2,-2\text{b}^2,-2\text{c}^2,$ are in A.P.
$\Rightarrow(\text{a}^2+\text{b}^2+\text{c}^2)-2\text{a},(\text{a}^2+\text{b}^2+\text{c}^2)-2\text{b}^2,\$\text{a}^2+\text{b}^2+\text{c}^2)-2\text{c}^2$ are in A.P.
$\Rightarrow(\text{b}^2+\text{c}^2-\text{a}^2),(\text{c}^2+\text{a}^2-\text{b}^2),(\text{b}^2+\text{a}^2-\text{c}^2)$ are in A.P.
$\Rightarrow\frac{(\text{b}^2+\text{c}^2-\text{a}^2)}{2\text{abc}},\frac{(\text{c}^2+\text{a}^2-\text{b}^2)}{2\text{abc}},\frac{(\text{b}^2+\text{a}^2-\text{c}^2)}{2\text{abc}}$ are in A.P.
$\Rightarrow\frac{1}{\text{a}}\frac{(\text{b}^2+\text{c}^2-\text{a}^2)}{2\text{bc}},\frac{1}{\text{b}}\frac{(\text{c}^2+\text{a}^2-\text{b}^2)}{2\text{ac}},\frac{1}{\text{c}}\frac{(\text{b}^2+\text{a}^2-\text{c}^2)}{2\text{ab}}$ are in A.P.
$\Rightarrow\frac{1}{\text{a}}\cos\text{A},\frac{1}{\text{b}}\cos\text{B},\frac{1}{\text{c}}\cos\text{C}$ are in A.P.
$\Rightarrow\frac{\text{k}}{\text{a}}\cos\text{A},\frac{\text{k}}{\text{b}}\cos\text{B},\frac{\text{k}}{\text{c}}\cos\text{C}$ are in A.P.
$\Rightarrow\frac{\cos\text{A}}{\sin\text{A}},\frac{\cos\text{B}}{\sin\text{B}},\frac{\cos\text{C}}{\sin\text{C}}$ are in A.P.
$\Rightarrow\cot\text{A},\cot\text{B},\cot\text{C}$ are in A.P.
View full question & answer→Question 85 Marks
$\text{a}(\cos\text{B}+\cos\text{C}-1)+\text{b}(\cos\text{C}+\cos\text{A}-1)\\+\text{c}(\cos\text{A}+\cos\text{B}-1)=0$
AnswerIn any $\triangle\text{ABC},$ we have
$\text{a = b}\cos\text{C + c}\cos\text{B}$
$\text{b = c}\cos\text{A + a}\cos\text{C}$
$\text{c = a}\cos\text{B + b}\cos\text{A}$
Therefore,
$\text{LHS}=\text{a}(\cos\text{B}+\cos\text{C}-1)+\text{b}(\cos\text{C}+\cos\text{A}-1)\\+\text{c}(\cos\text{A}+\cos\text{B}-1)$
$=\text{a}\cos\text{B + a}\cos\text{C}-\text{a + b}\cos\text{C + b}\cos\text{A}-\text{b + c}\cos\text{A + c}\cos\text{B}-\text{c}$
$=\text{c}-\text{b}\cos\text{A + a}\cos\text{C}-\text{a + a}-\text{c}\cos\text{B}\\+\text{b}\cos\text{A}-\text{b + b}-\text{a}\cos\text{C + c}\cos\text{B}-\text{c}$
$=0=\text{RHS}$
Hence proved.
View full question & answer→Question 95 Marks
$\frac{\sqrt{\sin\text{A}}-\sqrt{\sin\text{B}}}{\sqrt{\sin\text{A}}+\sqrt{\sin\text{B}}}=\frac{\text{a + b}-2\sqrt{\text{ab}}}{\text{a}-\text{b}}$
Answer$\frac{\sqrt{\sin\text{A}}-\sqrt{\sin\text{B}}}{\sqrt{\sin\text{A}}+\sqrt{\sin\text{B}}}=\frac{\text{a + b}-2\sqrt{\text{ab}}}{\text{a}-\text{b}}$
$\text{RHS}=\frac{\text{a + b}-2\sqrt{\text{ab}}}{\text{a}-\text{b}}$
$=\frac{(\sqrt{\text{a}})^2+(\sqrt{\text{b}})^2-2\sqrt{\text{ab}}}{(\sqrt{\text{a}})^2-(\sqrt{\text{b}})^2}$
$=\frac{(\sqrt{\text{a}}-\sqrt{\text{b}})^2}{(\sqrt{a})^2-(\sqrt{\text{b}})^2}$
$=\frac{(\sqrt{\text{a}}-\sqrt{\text{b}})}{(\sqrt{\text{a}}+\sqrt{\text{b}})}$
$=\frac{\big(\sqrt{\text{k}\sin\text{A}}-\sqrt{\text{k}\sin\text{B}}\big)}{\big(\sqrt{\text{k}\sin\text{A}}+\sqrt{\text{k}\sin\text{B}}\big)}$
$=\frac{\big(\sqrt{\sin\text{A}}-\sqrt{\sin\text{B}}\big)}{\big(\sqrt{\sin\text{A}}+\sqrt{\sin\text{B}}\big)}=\text{LHS}$ [taking k common and cancelling them]
Hence proved
View full question & answer→Question 105 Marks
$\frac{\text{a}^2\sin(\text{B}-\text{C})}{\sin\text{A}}+\frac{\text{b}^2\sin(\text{C}-\text{A})}{\sin\text{B}}+\frac{\text{c}^2\sin(\text{A}-\text{B})}{\sin\text{C}}=0$
Answer$\frac{\text{a}^2\sin(\text{B}-\text{C})}{\sin\text{A}}+\frac{\text{b}^2\sin(\text{C}-\text{A})}{\sin\text{B}}+\frac{\text{c}^2\sin(\text{A}-\text{B})}{\sin\text{C}}=0$
$\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\text{b}\sin\text{C}}=\frac{\text{c}}{\sin\text{C}}=\text{k}$
$\text{LHS}=\frac{\text{a}^2\sin(\text{B}-\text{C})}{\sin\text{A}}+\frac{\text{b}^2\sin(\text{C}-\text{A})}{\sin\text{B}}+\frac{\text{c}^2\sin(\text{A}-\text{B})}{\sin\text{C}}$
$=\text{ak}\sin(\text{B}-\text{C})+\text{bk}\sin(\text{C}-\text{A})+\text{ck}\sin(\text{A}-\text{B})$
$=\sin\text{A}\sin(\text{B}-\text{C})+\sin\text{B}\sin(\text{C}-\text{A})+\sin(\text{A}-\text{B})$
$=\sin(\pi-(\text{B + C}))\sin(\text{B}-\text{C})+\sin(\pi-(\text{C + A}))\\\sin(\text{C}-\text{A})+\sin(\pi-(\text{A + B}))\sin(\text{A}-\text{B})$
$=\sin(\text{B + C})\sin(\text{B}-\text{C})+\sin(\text{C + A})\\\sin(\text{C} - \text{A})+\sin(\text{A + B})\sin(\text{A}-\text{B})$
$=\sin^2\text{B}-\sin^2\text{C}+\sin^2\text{C}-\sin^2\text{A}+\sin^2\text{A}-\sin^2\text{B}=0=\text{RHS}$
View full question & answer→Question 115 Marks
$\text{a}\cos\text{A + b}\cos\text{B + c}\cos\text{C}=2\text{b}\sin\text{A}\sin\text{C}=2\text{c}\sin\text{A}\sin\text{B}$
AnswerLet $\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}=\text{k }...(1)$
Consider the LHS of the equation $\text{a}\cos\text{A}+\text{b}\cos\text{B}+\text{c}\cos\text{C}.$
$\text{a}\cos\text{A}+\text{b}\cos\text{B}+\text{c}\cos\text{C}$
$=\text{k}(\sin\text{A}\cos\text{A}+\sin\text{B}\cos\text{B}+\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}(2\sin\text{A}\cos\text{A}+2\sin\text{A}\cos\text{A}+2\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}(\sin2\text{A}+\sin2\text{B}+\sin2\text{C})$
$=\frac{\text{k}}{2}[2\sin(\text{A + B})\cos(\text{A}-\text{B})+2\sin\text{C}\cos\text{C}]$
$=\frac{\text{k}}{2}[2\sin(\pi - \text{C})\cos(\text{A}-\text{B})+2\sin\text{C}\cos\text{C}]$
$=\frac{\text{k}}{2}[2\sin\text{C}\cos(\text{A}-\text{B})+2\sin\text{C}\cos\text{C}]$
$=\frac{2\text{k}\sin\text{C}}{2}[\cos(\text{A}-\text{B})+\cos\text{C}]$
$=\text{k}\sin\text{C}[\cos(\text{A}-\text{B})+\cos\{\pi-(\text{A + B})\}]$
$=\text{k}\sin\text{C}[\cos(\text{A}-\text{B})-\cos(\text{A + B})]$
$=\text{k}\sin\text{C}[2\sin\text{A}\sin\text{B}]$
$=2\text{k}\sin\text{A}\sin\text{B}\sin\text{C }...(1)$
Now,
on putting $\text{k}\sin\text{C = C}$ in equation (1), we get:
$2\text{c}\sin\text{A}\sin\text{B}$
and on putting $\text{k}\sin\text{B = b}$ in equation (1), we get:
$2\text{b}\sin\text{A}\sin\text{C}$
So, from (1), we have
$\text{a}\cos\text{A + b}\cos\text{B + c}\cos\text{C}=2\text{b}\sin\text{A}\sin\text{C}=2\text{c}\sin\text{A}\sin\text{B}.$
Hence proved.
View full question & answer→Question 125 Marks
$\text{a}\sin\frac{\text{A}}{2}\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)+\text{b}\sin\frac{\text{B}}{2}\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)+\text{c}\sin\frac{\text{C}}{2}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)=0.$
Answer$\text{LHS}=\text{a}\sin\frac{\text{A}}{2}\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)+\text{b}\sin\frac{\text{B}}{2}\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)\\+\text{c}\sin\frac{\text{C}}{2}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$
We know $\text{a}\sin\text{B = b}\sin\text{A,c}\sin\text{B = b}\sin\text{C},\\\text{a}\sin\text{C}=\text{c}\sin\text{B}$
$\text{a}\sin\frac{\text{A}}{2}\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)+\text{b}\sin\frac{\text{B}}{2}\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)\\+\text{c}\sin\frac{\text{C}}{2}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)=0$
$=\text{a}\sin\Big(\frac{\pi-(\text{B + C})}{2}\Big)\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)+\text{b}\sin\Big(\frac{\pi-(\text{C + A})}{2}\Big)\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)\\+\text{c}\sin\Big(\frac{\pi-(\text{A + B})}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$
$=\text{a}\cos\Big(\frac{\text{B + C}}{2}\Big)\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)+\text{b}\cos\Big(\frac{\text{C + A}}{2}\Big)\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)\\+\text{c}\cos\Big(\frac{\text{A + B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$
$=\text{a}(\sin\text{B}-\sin\text{C})+\text{b}(\sin\text{C}-\sin\text{A})+\text{c}(\sin\text{A}-\sin\text{B})$
$=\text{a}\sin\text{B}-\text{a}\sin\text{C}+\text{b}\sin\text{C}-\text{b}\sin\text{A + c}\sin\text{A}-\text{c}\sin\text{B}$
$=\text{b}\sin\text{A}-\text{a}\sin\text{C + b}\sin\text{C}-\text{b}\sin\text{A + a}\sin\text{C}-\text{b}\sin\text{C}$
$=0 =\text{RHS}$
View full question & answer→Question 135 Marks
In a $\triangle\text{ABC},$ if $\cos\text{C}=\frac{\sin\text{A}}{2\sin\text{B}},$ prove that the triangle is isosceles.
AnswerLet $\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}=\frac{\sin\text{C}}{\text{c}}=\text{k}.$ Then, $\sin\text{A = ka},\sin\text{B = kb},\sin\text{C}=\text{kc}$
Now, $\cos\text{C}=\frac{\sin\text{A}}{2\sin\text{B}}$
$2\sin\text{B}\cos\text{C}=\sin\text{A}$
$2\Big(\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}\Big)\text{kb = ka}$
$\text{a}^2+\text{b}^2-\text{c}^2=\text{a}^2$
$\text{b}^2=\text{c}^2$
$\text{b = c}$
$\triangle\text{ABC}$ is isosceles.
View full question & answer→Question 145 Marks
In a $\triangle\text{ABC},$ prove that
$\sin^3\text{A}\cos(\text{B}-\text{C})+\sin^2\text{B}\cos(\text{C}-\text{A})+\sin^3\text{C}\cos(\text{A}-\text{B})\\=3\sin\text{A}\sin\text{B}\sin\text{C}$
AnswerLet $\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}=\text{k}...(1)$
$\text{LHS}=\sin^3\text{A}\cos(\text{B}-\text{C})+\sin^3\text{B}\cos(\text{C}-\text{A})+\sin^3\text{C}\cos(\text{A}-\text{B})$
$=\sin^2\text{A}\{\sin\text{A}\cos(\text{B}-\text{C})\}+\sin^2\text{B}\{\sin\text{B}\cos(\text{C}-\text{A})\}\\ \ \ \ \ \ \ \ +\sin^2\text{A}\{\sin\text{A}\cos(\text{A}-\text{B})\}$
$=\frac{\text{a}^2}{\text{k}^2}\{\sin\text{A}\cos(\text{B}-\text{C})\}+\frac{\text{b}^2}{\text{k}^2}\{\sin\text{B}\cos(\text{C}-\text{A})\}+\frac{\text{c}^2}{\text{k}^2}\{\sin\text{A}\cos(\text{A}-\text{B})\}$
$=\frac{1}{2\text{k}^2}\big[\text{a}^2\{2\sin\text{A}\cos(\text{B}-\text{C})\}+\text{b}^2\{2\sin\text{B}\cos(\text{C}-\text{A})\}\\ \ \ \ \ \ \ \ \ \ +\text{c}^2\{2\sin\text{A}\cos(\text{A}-\text{B})\}\big]$
$=\frac{1}{2\text{k}^2}\big[\text{a}^2\{2\sin(\pi-(\text{B + C}))\cos(\text{B} - \text{C})\}+\text{b}^2\{2\sin(\pi-(\text{A + C}))\cos(\text{C}-\text{A})\}\\ \ \ \ \ \ \ \ \ \ +\text{c}^2\{2\sin(\pi-(\text{B + C}))\cos(\text{A}-\text{B})\}\big]$
$=\frac{1}{2\text{k}^2}\big[\text{a}^2\{2\sin(\text{B + C})\cos(\text{B}-\text{C})\}+\text{b}^2\{2\sin(\text{C + A})\cos(\text{C}-\text{A})\}\\ \ \ \ \ \ \ \ \ \ \ +\text{c}^2\{2\sin(\text{A + B})\cos(\text{A}-\text{B})\}\big]$
$=\frac{1}{2\text{k}^2}\big[\text{a}^2\{\sin2\text{B}+\sin2\text{C}\}+\text{b}^2\{\sin2\text{C}+\sin2\text{A}\}+\text{c}^2\{\sin2\text{A}+\sin2\text{B}\}\big]$
$=\frac{1}{2\text{k}^2}\big[2\text{a}^2\{\sin\text{B}\cos\text{B}+\sin\text{C}\cos\text{C}\}+2\text{b}^2\{\sin\text{C}\cos\text{C}+\sin\text{A}\cos\text{A}\}\\ \ \ \ \ \ \ \ \ \ +2\text{c}^2\{\sin\text{A}\cos\text{A}+\sin\text{B}\cos\text{B}\}\big]$
$=\frac{1}{2\text{k}^3}\big[2\text{a}^2\{\text{k}\sin\text{B}\cos\text{B}+\text{k}\sin\text{C}\cos\text{C}\}+2\text{b}^2\{\text{k}\sin\text{C}\cos\text{C}+\text{k}\sin\text{A}\cos\text{A}\}\\ \ \ \ \ \ \ \ \ \ +2\text{c}^2\{\text{k}\sin\text{A}\cos\text{A}+\text{k}\sin\text{B}\cos\text{B}\}\big]$
$=\frac{1}{\text{k}^3}\big[\text{a}^2\{\text{b}\cos\text{B + c}\cos\text{C}\}+2\text{b}^2\{\text{c}\cos\text{C + a}\cos\text{A}\}\\ \ \ \ \ \ \ \ +2\text{c}^2\{\text{a}\cos\text{A + a}\cos\text{B}\}\big]$
$=\frac{1}{\text{k}^3}\big[\text{ab(a}\cos\text{B + b}\cos\text{A})+\text{bc(b}\cos\text{C + c}\cos\text{B})+\text{ac(a}\cos\text{ C + c}\cos\text{A})\big]$
$=\frac{1}{\text{k}^3}(\text{abc + bca + acb})$
$=3\text{abc}\times\frac{1}{\text{k}^3}$
$=3\sin\text{A}\sin\text{B}\sin\text{C}\times\frac{1}{\text{k}^3}\times\text{k}^3$
$=3\sin\text{A}\sin\text{B}\sin\text{C}$
$=\text{RHS}$
Hence proved.
View full question & answer→Question 155 Marks
$\text{a}(\cos\text{C}-\cos\text{B})=2(\text{b}-\text{c})\cos^2\frac{\text{A}}{2}.$
AnswerLet $\text{a = k}\sin\text{A}$
$\text{a}(\cos\text{C}-\cos\text{B})=2(\text{b}-\text{c})\cos^2\frac{\text{A}}{2}$
$\text{LHS}=\text{a}(\cos\text{C}-\cos\text{B})$
$=\text{a}2.\sin\frac{\text{C + B}}{2}.\sin\frac{\text{B}-\text{C}}{2}$
$=2\text{k}\sin\text{A}\sin\frac{\pi-\text{A}}{2}.\sin\frac{\text{B}-\text{C}}{2}$
$=2\text{k}2\sin\frac{\text{A}}{2}.\cos\frac{\text{A}}{2}.\cos\frac{\text{A}}{2}.\sin\frac{\text{B}-\text{C}}{2}$
$=2\text{k}\cos^2\frac{\text{A}}{2}\Big(2\sin\frac{\text{B}-\text{C}}{2}.\sin\frac{\text{A}}{2}\Big)$
$=2\text{k}\cos^2\frac{\text{A}}{2}\Big(2\sin\frac{\text{B}-\text{C}}{2}.\sin\frac{\pi-(\text{B + C})}{2}\Big)$
$=2\text{k}\cos^2\frac{\text{A}}{2}\Big(2\sin\frac{\text{B}-\text{C}}{2}.\cos\frac{\text{B + C}}{2}\Big)$
$=2\text{k}\cos^2\frac{\text{A}}{2}(\sin\text{B}-\sin\text{C})$
$=2\cos^2\frac{\text{A}}{2}(\text{k}\sin\text{B}-\text{k}\sin\text{C})$
$=2\cos^2\frac{A}{2}(\text{b}-\text{c})=\text{RHS}$
View full question & answer→Question 165 Marks
$\frac{\cos^2\text{B}-\cos^2\text{C}}{\text{b + c}}+\frac{\cos^2\text{C}-\cos^2\text{A}}{\text{c + a}}+\frac{\cos^2\text{A}-\cos^2\text{B}}{\text{a + b}}=0$
Answer$\frac{\cos^2\text{B}-\cos^2\text{C}}{\text{b + c}}+\frac{\cos^2\text{C}-\cos^2\text{A}}{\text{c + a}}+\frac{\cos^2\text{A}-\cos^2\text{B}}{\text{a + b}}=0$
$\text{LHS}=\frac{\cos^2\text{B}-\cos^2\text{C}}{\text{b + c}}+\frac{\cos^2\text{C}-\cos^2\text{A}}{\text{c + a}}+\frac{\cos^2\text{A}-\cos^2\text{B}}{\text{a + b}}$
$=\frac{\cos^2\text{B}-\cos^2\text{C}}{\text{b + c}}+\frac{\cos^2\text{C}-\cos^2\text{A}}{\text{c + a}}+\frac{\cos^2\text{A}-\cos^2\text{B}}{\text{a + b}}$
$=\frac{1-\sin^2\text{B}-1+\sin^2\text{C}}{\text{b + c}}+\frac{1-\sin^2\text{C}-1+\sin^2\text{A}}{\text{c + a}}+\frac{1-\sin^2\text{A}-1+\sin^2\text{B}}{\text{a +b}}$
$=\frac{\sin^2\text{C}-\sin^2\text{B}}{\text{b + c}}+\frac{\sin^2\text{A}-\sin^2\text{C}}{\text{c + a}}+\frac{\sin^2\text{B}-\sin^2\text{A}}{\text{a + b}}$
$=\frac{\text{k}^2\text{c}^2-\text{k}^2\text{b}^2}{\text{b + c}}+\frac{\text{k}^2\text{a}^2-\text{k}^2\text{c}^2}{\text{c + a}}+\frac{\text{k}^2\text{b}^2-\text{k}^2\text{a}^2}{\text{a + b}}$
$=\text{k}^2\Big(\frac{\text{c}^2-\text{b}^2}{\text{b + c}}+\frac{\text{a}^2-\text{c}^2}{\text{c + a}}+\frac{\text{b}^2-\text{a}^2}{\text{a + b}}\Big)$
$=\text{k}^2(\text{c}-\text{b + a}-\text{c + b}-\text{a})$ $[\text{Using }\text{b}^2 -\text{a}^2 = (\text{b}-\text{a})(\text{b + a})]$
$=0=\text{RHS}$
Hence Proved
View full question & answer→Question 175 Marks
$\frac{\text{c}}{\text{a}+\text{b}}=\frac{1-\tan\big(\frac{\text{A}}{2}\big)\tan\big(\frac{\text{B}}{2}\big)}{1+\tan\big(\frac{\text{A}}{2}\big)\tan\big(\frac{\text{B}}{2}\big)}$
Answer$\frac{\text{c}}{\text{a}+\text{b}}=\frac{1-\tan\big(\frac{\text{A}}{2}\big)\tan\big(\frac{\text{B}}{2}\big)}{1+\tan\big(\frac{\text{A}}{2}\big)\tan\big(\frac{\text{B}}{2}\big)}$
$\text{LHS}=\frac{\text{c}}{\text{a}+\text{b}}$
$=\frac{\text{k}\sin\text{C}}{\text{k}\sin\text{A}+\text{k}\sin\text{B}}$
$=\frac{2\sin\frac{\text{C}}{2}\cos\frac{\text{C}}{2}}{2\sin\big(\frac{\text{A +B}}{2}\big).\cos\big(\frac{\text{A}-\text{B}}{2}\big)}$
$=\frac{\sin\frac{\text{C}}{2}\cos\frac{\text{C}}{2}}{\sin\big(\frac{\pi-\text{C}}{2}\big).\cos\big(\frac{\text{A}-\text{B}}{2}\big)}$
$=\frac{\sin\big(\frac{\pi-\text{(A + B})}{2}\big)\cos\frac{\text{C}}{2}}{\cos\big(\frac{\text{C}}{2}\big).\cos\big(\frac{\text{A}-\text{B}}{2}\big)}$
$=\frac{\cos\big(\frac{\text{A + B}}{2}\big)}{\cos\big(\frac{\text{A}-\text{B}}{2}\big)}$
$=\frac{\cos\frac{\text{A}}{2}.\cos\frac{\text{B}}{2}-\sin\frac{\text{A}}{2}\sin\frac{\text{B}}{2}}{\cos\frac{\text{A}}{2}.\cos\frac{\text{B}}{2}+\sin\frac{\text{A}}{2}.\sin\frac{\text{B}}2{}}$
$=\frac{1-\tan\frac{\text{A}}{2}\tan\frac{\text{B}}2{}}{1+\tan\frac{\text{A}}{2}.\tan\frac{\text{B}}{2}}=\text{RHS}$ $\Big[$Dividing both Numberator and Denomiator by $\cos\big(\frac{\text{A}}{2}\big).\cos\big(\frac{\text{B}}{2}\big)\Big]$
View full question & answer→Question 185 Marks
In $\triangle\text{ABC}$ prove that, it $\theta$ be any angle, then $\text{b}\cos\theta=\text{c}\cos(\text{A}-\theta)+\text{a}\cos(\text{C}+\theta).$
Answer$\text{b}\cos\theta=\text{c}\cos(\text{A}-\theta)+\text{a}\cos(\text{C}+\theta)$
Let $\text{a}\sin\text{C = c}\sin\text{A}$ [Using sine rule]
$\text{RHS}=\text{c}\cos(\text{A}-\theta)+\text{a}\cos(\text{C}+\theta)$
$=\text{c}\cos\text{A}\cos\theta+\text{c}\sin\text{A}\cos\theta+\text{a}\cos\text{C}.\cos\theta-\text{a}\sin\text{C}\sin\theta$
$=\text{k}\sin\text{C}\cos\text{A}\cos\theta+\text{k}\sin\text{C}\sin\text{A}\cos\theta+\text{k}\sin\text{A}\cos\text{C}.\cos\theta\\-\text{k}\sin\text{A}\sin\text{C}\sin\theta$
$=\text{k}\sin\text{C}\cos\text{A}.\cos\theta+\text{k}\sin\text{A}\cos\text{C}.\cos\theta$
$=\text{k}\cos\theta(\sin\text{C}\cos\text{A}+\sin\text{A}\cos\text{C})$
$=\text{k}\cos\theta\sin(\text{C + A})$
$=\text{k}\cos\theta\sin(\pi-\text{B})$
$=\text{k}\cos\theta\sin\text{B}$
$=\text{k}\sin\text{B}.\cos\theta=\text{b}\cos\theta=\text{LHS}$
View full question & answer→Question 195 Marks
If the sides a, b, c of a $\triangle\text{ABC}$ are in H.P., prove that $\sin^2\frac{\text{A}}{2},\sin^2\frac{\text{B}}{2},\sin^2\frac{\text{C}}{2}$ are in H.P.
AnswerIf the sides a, b, c of a $\triangle\text{ABC}$ are in H.P.
$\therefore\frac{1}{\text{a}},\frac{1}{\text{b}}$ and $\frac{1}{\text{c}}$ are in AP
$\therefore\frac{1}{\text{b}}-\frac{1}{\text{a}}=\frac{1}{\text{c}}-\frac{1}{\text{b}}$
$\Rightarrow\frac{\text{a}-\text{b}}{\text{ba}}=\frac{\text{b}-\text{c}}{\text{ca}}$
$\Rightarrow\frac{\sin\text{A}-\sin\text{B}}{\sin\text{B}\sin\text{A}}=\frac{\sin\text{B}-\sin\text{C}}{\sin\text{C}\sin\text{B}}$ ...[Using sine rule]
$\Rightarrow\frac{2\sin\frac{\text{A}-\text{B}}{2}\cos\frac{\text{A + B}}{2}}{\sin\text{A}}=\frac{2\sin\frac{\text{B}-\text{C}}{2}\cos\frac{\text{B + C}}{2}}{\sin\text{C}}$
But $\text{A + B + C = } \pi$
$\text{A + B}=\pi-\text{C}$
$\cos\frac{\text{A + B}}{2}=\cos\Big(\frac{\pi}{2}-\frac{\text{C}}{2}\Big)=\sin\frac{\text{C}}{2}$
$\sin^2\frac{\text{C}}{2}\cos\frac{\text{C}}{2}\sin\frac{\text{A}-\text{B}}{2}=\sin\frac{\text{B}-\text{C}}{2}\cos\frac{\text{A}}{2}\sin^2\frac{\text{A}}{2}$
$\sin^2\frac{\text{C}}{2}\sin\frac{\text{A + B}}{2}\sin\frac{\text{A}-\text{B}}{2}=\sin\frac{\text{B}-\text{C}}{2}\cos\frac{\text{B + C}}{2}\sin^2\frac{\text{A}}{2}$
$\sin^2\frac{\text{C}}{2}\Big[\sin^2\frac{\text{A}}{2}-\sin^2\frac{\text{B}}{2}\Big]=\sin^2\frac{\text{A}}{2}\Big[\sin^2\frac{\text{B}}{2}-\sin^2\frac{\text{C}}{2}\Big]$
$\sin^2\frac{\text{C}}{2}\sin^2\frac{\text{A}}{2}-\sin^2\frac{\text{C}}{2}\sin^2\frac{\text{B}}{2}=\sin^2\frac{\text{A}}{2}\sin^2\frac{\text{B}}{2}-\sin^2\frac{\text{A}}{2}\sin^2\frac{\text{C}}{2}$
$\frac{1}{\sin^2\frac{\text{B}}{2}}-\frac{1}{\sin^2\frac{\text{A}}{2}}=\frac{1}{\sin^2\frac{\text{C}}{2}}-\frac{1}{\sin^2\frac{\text{B}}{2}}$
Hence $\frac{1}{\sin^2\frac{\text{A}}{2}},\frac{1}{\sin\frac{\text{B}}{2}},\frac{1}{\sin^2\frac{\text{C}}{2}}$ are in A.P.
$\therefore\sin^2\frac{\text{A}}{2},\sin^2\frac{\text{B}}{2},\sin^2\frac{\text{C}}{2}$ are in H.P.
View full question & answer→Question 205 Marks
In a $\triangle\text{ABC,}$ if $\sin^2\text{A}+\sin^2\text{B}=\sin^2\text{C},$ show that the triangle is right angled.
AnswerLet $\sin\text{A = ak},\sin\text{B = bk},\sin\text{C = ck}$
$\sin^2\text{A}+\sin^2\text{B}=\sin^2\text{C}$
$\Rightarrow{\text{k}}^2\text{a}^2+\text{k}^2\text{b}^2=\text{k}^2\text{c}^2$ [Using sine rule]
$\Rightarrow\text{a}^2+\text{b}^2=\text{c}^2$
Since the triangle satisfies the pythagoras theorem, therefore it is right angled.
View full question & answer→Question 215 Marks
A person observes the angle of elevation of the peak of a hill from a station to be $\alpha.$ He walks c metres along a slope inclined at an angle $\beta$ and finds the angle of elevation of the peak of the hill to be $\gamma.$ Show that the height of the peak above the ground is $\frac{\text{c}\sin\alpha\sin(\gamma-\beta)}{(\sin\gamma-\alpha)}.$
AnswerSuppose, AB is a peak whose height above the ground is t + x.
In $\triangle\text{DFC},$ $\sin\beta=\frac{\text{x}}{\text{c}}$ $\Rightarrow\text{x}=\text{c}\sin\beta$ and $\tan\beta=\frac{\text{x}}{\text{y}}$ $\Rightarrow\text{y}=\frac{\text{x}}{\tan\beta}=\frac{\text{c}\sin\beta}{\sin\beta}\times\cos\beta=\text{c}\cos\beta...(1)$ In $\triangle\text{ADE},$ $\tan\gamma=\frac{\text{t}}{\text{z}}$ $\Rightarrow\text{z}=\text{t}\cot\gamma...(2)$ In $\triangle\text{ABC},$ $\tan\text{a}=\frac{\text{t + x}}{\text{y + z}}$ $\Rightarrow\text{t + x}=(\text{c}\cos\beta\tan\alpha+\text{t}\cot\gamma)\tan\text{a}$ (from (1) and (2)) $\Rightarrow\text{t}-\text{t}\cot\gamma\tan\text{a = c}\cos\beta\tan\text{a}-\text{c}\sin\beta$ $(\because\text{x = c}\sin\beta)$ $\Rightarrow\text{t}\Big(1-\frac{\sin\alpha\cos\gamma}{\cos\alpha\sin\gamma}\Big)=\text{c}\Big(\frac{\cos\beta\sin\alpha-\cos\alpha\sin\beta}{\cos\alpha}\Big)$ $\Rightarrow\text{t}\Big(\frac{\sin\gamma\cos\alpha-\sin\alpha\cos\gamma}{\cos\alpha\sin\gamma}\Big)=\text{c}\frac{\sin(\alpha-\beta)}{\cos\alpha}$ $\Rightarrow\text{t}\frac{\sin(\gamma-\beta)}{\cos\alpha\sin\gamma}=\text{c}\frac{\sin(\alpha-\beta)}{\cos\alpha}$ $\Rightarrow\text{t = c}\frac{\sin\gamma\sin(\alpha-\beta)}{\sin(\gamma-\beta)}...(3)$ Now, $\text{AB = t + x = c}\frac{\sin\gamma\sin(\alpha-\beta)}{\sin(\gamma-\beta)}+\text{c}\sin\beta$ (using (3)) $=\text{c}\Big(\frac{\sin\gamma\sin(\alpha-\beta)}{\sin(\gamma-\beta)}+\sin\beta\Big)$ $=\text{c}\Big[\frac{\sin\gamma\sin(\alpha-\beta)+\sin(\gamma-\beta)}{\sin(\gamma-\beta)}\Big]$ $=\text{c}\Big[\frac{\sin\gamma\sin\alpha\cos\beta-\sin\beta\sin\gamma\cos\alpha+\sin\beta\sin\gamma\cos\alpha-\sin\beta\cos\gamma\sin\alpha}{\sin(\gamma-\beta)}\Big]$ $=\text{c}\Big[\frac{\sin\gamma\sin\alpha\cos\beta-\sin\beta\cos\gamma\sin\alpha}{\sin(\gamma-\beta)}\Big]$ $=\text{c}\Big[\frac{\sin\alpha\sin(\gamma-\beta)}{\sin(\gamma-\beta)}\Big]$ $=\frac{\text{c}\sin\alpha\sin(\gamma-\beta)}{\sin(\gamma-\beta)}$ Hence proved. View full question & answer→Question 225 Marks
$4\Big(\text{bc}\cos^2\frac{\text{A}}{2}+\text{ca}\cos^2\frac{\text{B}}{2}+\text{ab}\cos^2\frac{\text{C}}{2}\Big)=(\text{a + b + c})^2$
Answer$4\Big(\text{bc}\cos^2\frac{\text{A}}{2}+\text{ca}\cos^2\frac{\text{B}}{2}+\text{ab}\cos^2\frac{\text{C}}{2}\Big)=(\text{a + b + c})^2$$\text{LHS}=4\Big(\text{bc}\cos^2\frac{\text{A}}{2}+\text{ca}\cos^2\frac{\text{B}}{2}+\text{ab}\cos^2\frac{\text{C}}{2}\Big)$
$=2\Big(\text{bc.}2\cos^2\frac{\text{A}}{2}+\text{ca.}2\cos^2\frac{\text{B}}{2}+\text{ab}.2\cos^2\frac{\text{C}}{2}\Big)$
$=2(\text{bc.}(1-\cos\text{A})+\text{ca.}(1-\cos\text{B})+\text{ab.}(1-\cos\text{C}))$
$=2\text{bc}-2\text{bc}\cos\text{A}+2\text{ca}-2\text{ca}\cos\text{B}+2\text{ab}-2\text{ab}\cos\text{C}$
$=2\text{bc}-2\text{bc}\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}+2\text{ca}-2\text{ca}\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ca}}+2\text{ab}$
$-2\text{ab}\frac{\text{b}^2+\text{a}^2-\text{c}^2}{2\text{ab}}\text{[cos rule]}$
$=2\text{bc}-\text{b}^2-\text{c}^2+\text{a}^2+2\text{ca}-\text{a}^2-\text{c}^2+\text{b}^2+2\text{ab}-\text{b}^2-\text{a}^2+\text{c}^2$
$=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{ab}+2\text{ca}$
$=(\text{a + b + c})^2=\text{RHS}$
View full question & answer→Question 235 Marks
At the foot of a mountain, the elevation of it summit is 45°; after ascending 1000m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain.
Answer
$\text{DE}=1000\sin30=1000\times\frac{1}{2}=500\text{m}=\text{FB}$
$\text{EC}=1000\cos30=1000\times\frac{\sqrt{3}}{2}=500\sqrt{3}\text{m}$
Let AF = xm
$\text{DF}=\frac{\text{x}}{\sqrt{3}}\text{m = BE}$
WE know,
From $\triangle\text{ABC},$
$\tan45=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow1=\frac{\text{AF + FB}}{\text{BE + EC}}$
$\Rightarrow1=\frac{\text{x}+500}{\frac{\text{x}}{\sqrt{3}}+500\sqrt{3}}$
$\Rightarrow\frac{\text{x}}{\sqrt{3}}+500\sqrt{3}=\text{x}+500$
$\Rightarrow\text{x}+1500=\text{x}\sqrt{3}+500\sqrt{3}$
$\Rightarrow1500-500\sqrt{3}=\text{x}\sqrt{3}-\text{x}$
$\Rightarrow500\sqrt{3}(\sqrt{3}-1)=\text{x}(\sqrt{3}-1)$
$\therefore\text{x}=500\sqrt{3}\text{m}$
The height of the mountain is $\text{AB = AF + FB}=500(\sqrt{3}+1)\text{m}$ View full question & answer→Question 245 Marks
Two ships leave a port at the same time. One goes 24km/ hr in the direction N 38° E and other travels 32km/ hr in the direction S 52° E. Find the distance between the ships at the end of 3hrs.
Answer
Let P and Q be the position of two ships at the end of 3 hours.
Then,
$\text{OP}=3\times24=72\text{km}$ and $\text{OQ}=3\times32=96\text{km}$
Using cosine formula in $\triangle\text{OPQ},$ we get
$\text{PQ}^2=\text{OP}^2+\text{OQ}^2-2\text{OP}\times\text{OQ}\cos90^{\circ}$
$\text{PQ}^2=72^2+96^2-2\times72\times96\cos90^{\circ}$
$\text{PQ}^2=14400$
$\text{PQ}=120\text{km}$ View full question & answer→Question 255 Marks
If in a $\triangle\text{ABC},\cos^2\text{A}+\cos^2\text{B}+\cos^2\text{C}=1,$ prove that the triangle is right angled.
AnswerLet ABC be any triangle.
In $\triangle\text{ABC},$
$\cos^2\text{A}+\cos^2\text{B}+\cos^2\text{C}=1$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}+\cos^2[\pi-(\text{B + A})]=1$ $(\because\text{A + B + C} = \pi)$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}+\cos^2(\text{B + A})=1$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}=1-\cos^2(\text{B + A})$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}=\sin^2(\text{B + A})$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}=(\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B})^2$
$\Rightarrow\cos^2\text{A}+\cos^2\text{B}=\sin^2\text{A}\cos^2\text{B}+\cos^2\text{A}\sin^2\text{B}\\+2\sin\text{A}\sin\text{B}\cos\text{A}\cos\text{B}$
$\Rightarrow\cos^2\text{A}(1-\sin^2\text{B})+\cos^2\text{B}(1-\sin^2\text{A})$ $=2\sin\text{A}\sin\text{B}\cos\text{A}\cos\text{B}$
$\Rightarrow2\cos^2\text{A}\cos^2\text{B}=2\sin\text{A}\sin\text{B}\cos\text{A}\cos\text{B}$
$\Rightarrow\cos\text{A}\cos\text{B}=\sin\text{A}\sin\text{B}$
$\Rightarrow\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}=0$
$\Rightarrow\cos(\text{A + B)}=0$
$\Rightarrow\cos\text{(A + B)}=\cos90^{\circ}$
$\Rightarrow\text{A + B}=90^{\circ}$
$\Rightarrow\text{C}=90^{\circ}$ $(\because\text{A + B + C }= 180^{\circ})$
Hence, $\triangle\text{ABC}$ is right angled.
View full question & answer→Question 265 Marks
$\text{a}(\cos\text{B}\cos\text{C}+\cos\text{A})=\text{b}(\cos\text{C}\cos\text{A}+\cos\text{B})\\=\text{c}(\cos\text{A}\cos\text{B}+\cos\text{C}).$
AnswerSuppose $\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}=\text{k}$Consider:
$\text{a}(\cos\text{B}\cos\text{C}+\cos\text{A})$
$=\text{k}\sin\text{A}(\cos\text{B}\cos\text{C}+\cos\text{A})$
$=\text{k}(\sin\text{A}\cos\text{B}\cos\text{C}+\cos\text{A}\sin\text{A})$
$=\text{k}\Big[\frac{1}{2}\cos\text{C}\{\sin(\text{A + B})+\sin(\text{A}-\text{B})\}+\sin\text{A}\cos\text{A}\Big]$
$=\text{k}\Big[\frac{1}{2}\{\sin(\text{A + B})\cos\text{C}+\sin(\text{A}-\text{B})\cos\text{C}\}+\sin\text{A}\cos\text{A}\Big]$
$=\text{k}\Big[\frac{1}{2}\big\{\frac{1}{2}\big[\sin(\text{A + B + C})+\sin(\text{A + B}-\text{C})\\+\sin(\text{A}-\text{B + C})+\sin(\text{A}-\text{B}-\text{C})\big]\big\}+\sin\text{A}\cos\text{A}]$
$=\text{k}\Big[\frac{1}{4}\{\sin\pi+\sin(\pi-2\text{C})+\sin(\pi-2\text{B})-\sin(\pi-2\text{A})\}+\frac{\sin2\text{A}}{2}\Big]$ $(\because\text{A + B + C} = \pi)$
$=\frac{\text{k}}{4}(\sin2\text{C}+\sin2\text{B}+\sin2\text{A})...(1)$
and
$\text{b}(\cos\text{A}\cos\text{C}+\cos\text{B})$
$=\text{k}(\sin\text{B}\cos\text{A}\cos\text{C}+\sin\text{B}\cos\text{B})$
$=\text{k}\Big[\frac{1}{2}\cos\text{A}\{\sin(\text{B + C})+\sin(\text{B} -\text{C})\}+\frac{\sin2\text{B}}{2}\Big]$
$=\text{k}\Big(\frac{1}{2}(\sin(\text{B + C})\cos\text{A}+\sin(\text{B}-\text{C})\cos\text{A})+\frac{\sin2\text{B}}{2}\Big)$
$=\text{k}\Big(\frac{1}{4}(\sin(\text{B + C + A})+\sin(\text{B + C}-\text{A})\\+\sin(\text{B}-\text{C + A})+\sin(\text{B}-\text{C}-\text{A}))+\frac{\sin2\text{B}}{2}\Big)$
$=\frac{\text{k}}{2}\Big(\sin\pi+\sin(\pi-2\text{A})+\sin(\pi-2\text{C})-\sin(\pi-2\text{B})+\frac{\sin2\text{B}}{2}\Big)$ $(\because\text{A + B + C}=\pi)$
$=\frac{\text{k}}{4}(\sin2\text{A}+\sin2\text{C}+\sin2\text{B})...(2)$
Similarly,
$\text{c}(\cos\text{A}\cos\text{B}+\cos\text{C})=\frac{\text{k}}{4}(\sin2\text{A}+\sin2\text{B}+\sin2\text{C})...(3)$
From (1), (2) and (3), we get:
$\text{a}(\cos\text{B}\cos\text{C}+\cos\text{A})=\text{b}(\cos\text{C}\cos\text{A}+\cos\text{B})\\=\text{c}(\cos\text{A}\cos\text{B}+\cos\text{C})$
Hence proved.
View full question & answer→Question 275 Marks
In a $\triangle\text{ABC},$ if $\angle\text{B}=60^{\circ},$ prove that (a + b + c) = 3ca
AnswerGiven, $\angle\text{B}=60^{\circ}$
We know that, $\cos\text{B}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}$
$\Rightarrow\cos60^{\circ}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}$
$\Rightarrow\frac{1}{2}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}$ $\Big(\because\cos60^{\circ}=\frac{1}{2}\Big)$
$\Rightarrow\text{ac}=\text{a}^2+\text{c}^2-\text{b}^2$
$\Rightarrow3\text{ac}-2\text{ac}=\text{a}^2+\text{c}^2-\text{b}^2$
$\Rightarrow3\text{ac}=\text{a}^2+\text{c}^2-\text{b}^2+2\text{ac}$
$\Rightarrow3\text{ac}=\text{a}^2+\text{c}^2+2\text{ac}-\text{b}^2$
$\Rightarrow3\text{ac}=(\text{a + c})^2-\text{b}^2$
$\Rightarrow3\text{ac}=(\text{a + c + b})\text{(a + c}-\text{b})$
$\Rightarrow3\text{ac}=(\text{a + b + c})(\text{a}-\text{b + c})$
Hence proved.
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