(Each question 2 marks)

Question 12 Marks

In a $\triangle\text{ABC},$ if a = 5, b = 6 and C = 60°, show that its area is $\frac{15\sqrt{3}}{2}\text{ sq. units.}$

Answer

The area of a triangle ABC is given by $\triangle=\frac{1}{2}\text{ab}\sin\text{C}$ $=\frac{1}{2}\times5\times6\sin60^{\circ}$ $=\frac{15\sqrt{3}}{2}\text{sq. unit}$

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Question 22 Marks

If in any $\triangle\text{ABC},\angle\text{C}=105^\circ,\angle\text{B}=45^{\circ},\text{a}=2,$ then find b.

Answer

$\angle\text{C}=105^{\circ},\angle\text{B}=45^{\circ},\text{a}=2$ From here we can calculate that $\angle{\text{A}}=30^{\circ}$ $\text{a}\sin\text{B}=\text{b}\sin\text{A}$ $\Rightarrow2\sin45=\text{b}\sin30$ $\Rightarrow{2}\times\frac{1}{\sqrt{2}}=\text{b}\sin30$ $\Rightarrow2\times\frac{1}{\sqrt{2}}=\text{b}\times\frac{1}{2}$ $\Rightarrow{}\sqrt{2}=\frac{\text{b}}{2}$ $\Rightarrow\text{b}=2\sqrt{2}$

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Question 32 Marks

$(\text{a}-\text{b})\cos\frac{\text{C}}{2}=\text{c}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$

Answer

$(\text{a}-\text{b})\cos\frac{\text{C}}{2}=\text{c}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$ Let $\text{a = k}\sin\text{A, b = k}\sin\text{B,c = k}\sin\text{C}$ $\text{LHS}=\text{(a}-\text{b})\cos\frac{\text{C}}{2}$ $=\text{k}(\sin\text{A}-\sin\text{B}).\cos\frac{\text{C}}{2}$ $=2\text{k}\cos\Big(\frac{\text{A + B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big).\cos\frac{\text{C}}{2}$ $=2\text{k}\cos\Big(\frac{\pi-\text{C}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big).\cos\frac{\text{C}}2{}$ $=2\text{k}\sin\Big(\frac{\text{C}}{2}\Big).\cos\frac{\text{C}}2{}.\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$ $\Big[\cos\Big(\frac{\pi}{2}-\theta\Big)=\sin\theta\Big]$ $=\text{k}\sin\text{C}.\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$ $=\text{c}.\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)=\text{RHS}$

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Question 42 Marks

$\text{a}(\sin\text{B}-\sin\text{C})+\text{b}(\sin\text{C}-\sin\text{A})+\text{c}(\sin\text{A}-\sin\text{B})=0$

Answer

$\text{a}(\sin\text{B}-\sin\text{C})+\text{b}(\sin\text{C}-\sin\text{A})+\text{c}(\sin\text{A}-\sin\text{B})=0$ $\text{LHS}=\text{a}(\sin\text{B}-\sin\text{C})+\text{b}(\sin\text{C}-\sin\text{A})+\text{c}(\sin\text{A}-\sin\text{B})$ $=\text{a}\sin\text{B}-\text{a}\sin\text{C}+\text{b}\sin\text{C}-\text{b}\sin\text{A}+\text{c}\sin\text{A}-\text{c}\sin\text{B}$ $=\text{b}\sin\text{A}-\text{c}\sin\text{A + c}\sin\text{B}-\text{b}\sin\text{A + c}\sin\text{A}-\text{c}\sin\text{B}$ $[\because \text{b}\sin\text{A = a}\sin\text{B,b}\sin\text{C = c}\sin]$ $=0=\text{RHS}$ Hence Proved

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Question 52 Marks

$\frac{\cos2\text{A}}{\text{a}^2}-\frac{\cos2\text{B}}{\text{b}^2}=\frac{1}{\text{a}^2}-\frac{1}{\text{b}^2}$

Answer

$\text{LHS}=\frac{\cos2\text{A}}{\text{a}^2}-\frac{\cos2\text{B}}{\text{b}^2}$ $\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}=\frac{\sin\text{C}}{\text{c}}=\text{k}$ $=\frac{1-2\sin^2\text{A}}{\text{a}^2}-\frac{1-2\sin^2\text{B}}{\text{b}^2}$ $=\frac{1}{\text{a}^2}-\frac{1}{\text{b}^2}-2\Big(\frac{\sin^2\text{A}}{\text{a}^2}-\frac{\sin^2\text{B}}{\text{b}^2}\Big)$ $=\frac{1}{\text{a}^2}-\frac{1}{\text{b}^2}-2(\text{k}^2-\text{k}^2)$ [Using sine rule] $=\frac{1}{\text{a}^2}-\frac{1}{\text{b}^2}=\text{RHS}$ Hence Proved

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Question 62 Marks

In triangle ABC, prove the following: $\frac{\text{a}-\text{b}}{\text{a + b}}=\frac{\tan\big(\frac{\text{A}-\text{B}}{2}\big)}{\tan\big(\frac{\text{A}+\text{B}}{2}\big)}$

Answer

$\frac{\text{a}-\text{b}}{\text{a + b}}=\frac{\tan\big(\frac{\text{A}-\text{B}}{2}\big)}{\tan\big(\frac{\text{A}+\text{B}}{2}\big)}$ Let $\text{a = k}\sin\text{A,b = k}\sin\text{B}$ (Using sine rule) $\text{LHS}=\frac{\text{a}-\text{b}}{\text{a + b}}$ $=\frac{\text{k}\sin\text{A}-\text{k}\sin\text{B}}{\text{k}\sin\text{A}+\text{k}\sin\text{B}}$ $=\frac{\sin\text{A}-\sin\text{B}}{\sin\text{A}+\sin\text{B}}$ $=\frac{2\cos\big(\frac{\text{A +B}}{2}\big)\sin\big(\frac{\text{A}-\text{B}}{2}\big)}{2\sin\big(\frac{\text{A +B}}{2}\big)\cos\big(\frac{\text{A}-\text{B}}{2}\big)}$ $=\frac{\tan\big(\frac{\text{A}-\text{B}}{2}\big)}{\tan\big(\frac{\text{A}+\text{B}}{2}\big)}=\text{RHS}$

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Question 72 Marks

If in a $\triangle\text{ABC},\angle\text{A}=45^\circ,\angle\text{B}=60^{\circ},$ and $\angle\text{C}=75^{\circ},$ find the ratio of its sides.

Answer

$\angle\text{A}=45^{\circ},\angle\text{B}=60^\circ$ and $\angle\text{C}=75^\circ$ Using sing rule, $\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}=\text{k}$ $\frac{\text{a}}{\sin\text{45}}=\frac{\text{b}}{\sin\text{60}}=\frac{\text{c}}{\sin\text{75}}=\text{k}$ $\frac{\text{a}}{\frac{1}{\sqrt{2}}}=\frac{\text{b}}{\frac{\sqrt{3}}{2}}=\frac{\text{c}}{\frac{\sqrt{3}+1}{2\sqrt{2}}}=\text{k}$ $\text{a:b:c}=2:\sqrt{6}:(\sqrt{3}+1)$

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Question 82 Marks

In $\triangle\text{ABC},$ if a= 18, b = 24 and c = 30 and $\angle\text{C}=90^{\circ},$ find $\sin\text{A},\sin\text{B}$ and $\sin\text{C}.$

Answer

a = 18, b = 24, c = 30, $\angle\text{C}=90^{\circ}$ let $\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}=\frac{\sin\text{C}}{\text{c}}$ $\frac{\sin\text{A}}{\text{18}}=\frac{\sin\text{B}}{\text{24}}=\frac{\sin\text{90}}{\text{30}}$ $\frac{\sin\text{A}}{18}=\frac{\sin\text{B}}{24}=\frac{1}{30}$ $\frac{\sin\text{A}}{18}=\frac{1}{30}\Rightarrow\sin\text{A}=\frac{18}{30}=\frac{3}{5}$ $\frac{\sin\text{B}}{24}=\frac{1}{30}\Rightarrow\sin\text{B}=\frac{24}{30}=\frac{4}{5}$ $\therefore\sin\text{}\text{A}=\frac{3}{5},\sin\text{B}=\frac{4}{5},\sin\text{C}=1$

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Question 92 Marks

The sides of a triangle are a = 5, b = 6 and c = 8, show that: $8\cos\text{A}+16 \cos\text{B}+4\cos\text{C}=17.$

Answer

We have, a = 4, b = 6 and c = 8 $\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{ab}}=\frac{7}{8}$ $\cos\text{B}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ab}}=\frac{11}{16}$ $\cos\text{C}=\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}=-\frac{1}{4}$ $8\cos\text{A}+16\cos\text{B}+4\cos\text{C}=8\times\frac{7}{8}+16\times\frac{11}{16}+4\times\Big(-\frac{1}{4}\Big)$ $=17$

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Question 102 Marks

In a $\triangle\text{ABC},$ if $\text{a}=\sqrt{2},\text{b}=\sqrt{3}$ and $\text{c}=\sqrt{5},$ show that its area is $\frac{1}{2}\sqrt{6 }\text{ sq. units.}$

Answer

The area of a triangle ABC is given by $\triangle=\frac{1}{2}\text{ab}\sin\text{C}$ $\cos\text{C}=\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}$ $=\frac{2+3-5}{2\sqrt{6}}$ $=0$ $\sin\text{C}=\sqrt{1-\cos^2\text{C}}$ $=1$ Therefore, $\triangle=\frac{1}{2}\text{ab}\sin\text{C}$ $\frac{1}{2}\sqrt{6}$

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