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Question 11 Mark

In any triangle ABC, find the value of $\text{a}\sin(\text{B}-\text{C})+\text{b}\sin(\text{C}-\text{A})+\text{c}\sin(\text{A}-\text{B}).$

Answer

$\text{a}\sin(\text{B}-\text{C})+\text{b}\sin(\text{C}-\text{A})+\text{c}\sin(\text{A}-\text{B})$
$=\text{k}\sin\text{A}\sin(\text{B}-\text{C})+\text{k}\sin\text{B}\sin(\text{C}-\text{A})+\text{k}\sin\text{C}\sin(\text{A}-\text{B})$
$=\text{k}\sin(\pi-(\text{B + C}))\sin(\text{B}-\text{C})+\text{k}\sin(\pi-(\text{C + A}))\sin(\text{C}-\text{A})\\+\text{k}\sin(\pi-(\text{A + B}))\sin(\text{A}-\text{B})$
$=\text{k}\sin(\text{B + C})\sin(\text{B}-\text{C})+\text{k}\sin(\text{C + A})\\\sin(\text{C}-\text{A})+\text{k}\sin(\text{A + B})\sin(\text{A}-\text{B})$
$=\text{k}\big[\sin^2\text{B}-\sin^2\text{C}+\sin^2\text{C}-\sin^2\text{A}+\sin^2\text{A}-\sin^2\text{B}\big]$
$=\text{k}\times0 = 0$

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Question 21 Mark

In a $\triangle\text{ABC},$ if $\text{b}=\sqrt{3},$ c = 1 and $\angle\text{A}=30^{\circ},$ find a.

Answer

We know,
$\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}$
$\Rightarrow\cos30=\frac{\sqrt{3^2}+1^2-\text{a}^2}{2\times\sqrt{3}\times1}$
$\Rightarrow\frac{\sqrt{3}}{2}=\frac{3+1-\text{a}^2}{2\sqrt{3}}$
$\Rightarrow2\sqrt{3}\times\sqrt{3}=2(4-\text{a}^2)$
$\Rightarrow3=4-\text{a}^2$
$\Rightarrow\text{a}^2=1$
$\Rightarrow\text{a}=1$

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Question 31 Mark

If in a $\triangle\text{ABC},\frac{\cos\text{A}}{\text{a}}=\frac{\cos\text{B}}{\text{b}}=\frac{\cos\text{C}}{\text{c}},$ then find the measures of angles A, B, C.

Answer

$\frac{\cos\text{A}}{\text{a}}=\frac{\cos\text{B}}{\text{b}}=\frac{\cos\text{C}}{\text{c}}$
$\Rightarrow\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{abc}}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{abc}}=\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{abc}}$
$\Rightarrow\text{b}^2+\text{c}^2-\text{a}^2=\text{a}^2+\text{c}^2-\text{b}^2=\text{a}^2+\text{b}^2-\text{c}^2$
$\therefore\text{b}^2+\text{c}^2-\text{a}^2=\text{a}^2+\text{c}^2-\text{b}^2$
$2\text{b}^2=2\text{a}^2\Rightarrow\text{b = a}$
$\therefore\text{a}^2+\text{c}^2-\text{b}^2=\text{a}^2+\text{b}^2-\text{c}^2$
$2\text{c}^2=2\text{b}^2\Rightarrow\text{c = b}$
$\text{b}^2+\text{c}^2-\text{a}^2=\text{a}^2+\text{b}^2-\text{c}^2$
$\Rightarrow2\text{c}^2=2\text{a}^2\Rightarrow\text{c = a}$
Hence $\text{a = b = c}$
or, $\text{A = B = C}=60^{\circ}$

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Question 41 Mark

In a $\triangle\text{ABC},$ if $\sin\text{A}$ and $\sin\text{B}$ are the roots of the equation $\text{c}^2\text{x}^2-\text{c(a + b) x + ab}=0,$ then find $\angle\text{C}.$

Answer

$\text{c}^2\text{x}^2-\text{c(a + b) x + ab}=0$
$\sin\text{A }\&\sin\text{B}$ are the roots of the equation,
$\sin\text{A}+\sin\text{B}=\frac{\text{c}(\text{a + b})}{\text{c}^2}=\frac{(\text{a} + \text{b})}{\text{c}}$
$\sin\text{A}\sin\text{B}=\frac{\text{ab}}{\text{c}^2}$
$\sin\text{A}-\sin\text{B}=\sqrt{(\sin\text{A}+\sin\text{B})^2-4\sin\text{A}.\sin\text{B}}$
$=\sqrt{\bigg(\frac{\text{(a + b)}}{\text{c}}\bigg)^2-4\frac{\text{ab}}{\text{c}^2}}$
$=\sqrt{\frac{\text{a}^2+\text{b}^2+2\text{ab}-4\text{ab}}{\text{c}^2}}$
$=\sqrt{\frac{\text{a}^2+\text{b}^2-2\text{ab}}{\text{c}^2}}$
$=\sqrt{\frac{(\text{a}-\text{b})^2}{\text{c}^2}}=\frac{\text{(a}-\text{b})}{\text{c}}$
$\sin\text{A}+\sin\text{B}=\frac{(\text{a + b})}{\text{c}}...(\text{i})$
$\sin\text{A}-\sin\text{B}=\frac{(\text{a}-\text {b})}{\text{c}}...(\text{ii})$
Adding (i) and (ii), we get,
$\sin\text{A}=\frac{\text{a}}{\text{c}}$ and $\sin\text{B}=\frac{\text{b}}{\text{c}}$
$\sin\text{A}=\frac{\text{prependicular}}{\text{hypotenuse}}$
So, in $\triangle\text{ABC},$
one of the side = a, another side = b, third side = c
$\therefore\text{a}^2+\text{b}^2=\text{c}^2$
$\angle\text{C}=90^{\circ}$

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Question 51 Mark

Find the area of the triangle $\triangle\text{ABC}$ in which a = 1, b = 2 and $\angle\text{C}=60^{\circ}.$

Answer

Area of the $\triangle\text{ABC}=\frac{1}{2}\sin\text{C}$
$=\frac{1}{2}\times1\times2\times\sin60$
$=\frac{\sqrt{3}}2{}\text{sq. units}$

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Question 61 Mark

In any $\triangle\text{ABC},$ find the value of $\sum\text{a}(\sin\text{B}-\sin\text{C})$

Answer

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

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Question 71 Mark

In a $\triangle\text{ABC},$ if $\cos\text{A}=\frac{\sin\text{B}}{2\sin\text{C}'}$ then show that c = a.

Answer

Let $\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}=\frac{\sin\text{C}}{\text{c}}=\text{k}$
$\cos\text{A}=\frac{\sin\text{B}}{2\sin\text{C}}$
$\Rightarrow\Big(\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}\Big)=\frac{\text{kb}}{2\text{kc}}$
$\Rightarrow\Big(\frac{\text{b}^2+\text{c}^2-\text{a}^2}{\text{b}}\Big)=\text{b}$
$\Rightarrow\text{b}^2+\text{c}^2-\text{a}^2=\text{b}^2$
$\Rightarrow\text{c}^2=\text{a}^2$
$\Rightarrow\text{c = a}$

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Question 81 Mark

In a $\triangle\text{ABC},$ if b = 20, c = 21 and $\sin\text{A}=\frac{3}{5},$ find a.

Answer

$\text{b = 20, c = 21,}\sin\text{A}=\frac{3}{5}$
$\cos\text{A}=\frac{4}{5}$
$\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}$
$\Rightarrow\frac{4}{5}=\frac{20^2+21^2-\text{a}^2}{2\times20\times21}$
$\Rightarrow2\times4\times4\times21=841-\text{a}^2$
$\Rightarrow672=841-\text{a}^2$
$\Rightarrow\text{a}^2=841-672$
$\Rightarrow\text{a}^2=169$
$\Rightarrow\text{a}=13$

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Question 91 Mark

In a $\triangle\text{ABC},$ if a = 8, b = 10, c = 12 and $\text{C}=\lambda\text{A},$ find the value of $\lambda.$

Answer

Using cosine rule, we have
$\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}$
$\Rightarrow\cos\text{A}=\frac{10^2+12^2-8^2}{2\times10\times12}$
$\Rightarrow\cos\text{A}=\frac{100+144-64}{240}$
$\Rightarrow\cos\text{A}=\frac{180}{240}=\frac{3}{4}...(1)$
Now, using sine rule, we have
$\Rightarrow\frac{\text{a}}{\sin\text{A}}=\frac{\text{c}}{\sin\text{C}}$
$\Rightarrow\frac{8}{\sin\text{A}}=\frac{12}{\sin\lambda\text{A}}$
$\Rightarrow\sin\lambda\text{A}=\frac{3}{2}\sin\text{A}$
$\Rightarrow\sin\lambda\text{A}=2\times\frac{3}{4}\sin\text{A}$
$\Rightarrow\sin\lambda\text{A}=2\sin\text{A}\cos\text{A}$ [Using (1)]
$\Rightarrow\sin\lambda\text{A}=\sin2\text{A}$
$\Rightarrow\lambda=2$

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Question 101 Mark

If the sides of a triangle are proportional to 2, $\sqrt{6}$ and $\sqrt{3}-1,$ find the measure of its greatest angle.

Answer

let $\text{a}=2,\text{b}=\sqrt{6},\text{c}=\sqrt{3}-1$
$\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}$
$\Rightarrow\cos\text{A}=\frac{6+(3+1-2\sqrt{3})-4}{2\times\sqrt{6}\times(\sqrt{3}-1)}$
$\Rightarrow\cos\text{A}=\frac{6-2\sqrt{3}}{2\times\sqrt{6}\times(\sqrt{3}-1)}$
$\Rightarrow\cos\text{A}=\frac{2\sqrt{3}(\sqrt{3}-1)}{2\sqrt{6}(\sqrt{3}-1)}$
$\Rightarrow\cos\text{A}=\frac{1}{\sqrt{2}}$
$\text{A}=45$
$\cos\text{C}=\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}$
$\Rightarrow\cos\text{A}=\frac{4+6-(3+1-2\sqrt{3})}{2\times2\times\sqrt{6}}$
$\Rightarrow\cos\text{A}=\frac{6+2\sqrt{3}}{2\times2\times\sqrt{6}}$
$\Rightarrow\cos\text{A}=\frac{2\sqrt{3}(\sqrt{3}+1)}{4\sqrt{6}}$
$\Rightarrow\cos\text{A}=\frac{(\sqrt{3}+1)}{2\sqrt{2}}$
$\text{A}=15$
$\angle\text{C}=180-(\text{A + B})$
$=180-(45+15)$
$=120^{\circ}$

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