Exercise 3.9 - Chapter 3 Trigonometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 3.9
Question 1.
In a $\triangle \mathrm{ABC}$, if $\frac{\sin \mathrm{A}}{\sin \mathrm{C}}=\frac{\sin (\mathrm{A}-\mathrm{B})}{\sin (\mathrm{B}-\mathrm{C})}$, prove that $a^2, b^2, c^2$ are in Arithmetic Progression.
Solution:
$
\begin{aligned}
\mathrm{LHS}=\frac{\sin \mathrm{A}}{\sin \mathrm{C}} & =\frac{\frac{a}{2 \mathrm{R}}}{\frac{c}{2 \mathrm{R}}}=\frac{a}{c} \\
\mathrm{RHS} & =\frac{\sin (\mathrm{A}-\mathrm{B})}{\sin (\mathrm{B}-\mathrm{C})}=\frac{\sin \mathrm{A} \cos \mathrm{B}-\cos \mathrm{A} \sin \mathrm{B}}{\sin \mathrm{B} \cos \mathrm{C}-\cos \mathrm{B} \sin \mathrm{C}} \\
\mathrm{Nr} & =\frac{a}{2 \mathrm{R}}\left[\frac{a^2+c^2-b^2}{2 a c}\right]-\frac{b}{2 \mathrm{R}}\left[\frac{b^2+c^2-a^2}{2 b c}\right] \\
& =\frac{1}{2 \mathrm{R}}\left\{\frac{a^2+c^2-b^2}{2 c}-\frac{b^2+c^2-a^2}{2 c}\right\} \\
& =\frac{1}{2 \mathrm{R}}\left[\frac{a^2+c^2-b^2-b^2-c^2+a^2}{2 c}\right] \\
& =\frac{1}{2 \mathrm{R}}\left[\frac{2 a^2-2 b^2}{2 a b}\right]=\frac{1}{2 \mathrm{R}}\left[\frac{a^2-b^2}{c}\right]
\end{aligned}
$

$
\begin{aligned}
& \operatorname{Dr}=\frac{b}{2 \mathrm{R}}\left[\frac{a^2+b^2-c^2}{2 a b}\right]-\frac{c}{2 \mathrm{R}}\left[\frac{a^2+c^2-b^2}{2 a c}\right] \\
& =\frac{1}{2 R}\left[\frac{a^2+b^2-c^2}{2 a}-\frac{a^2+c^2-b^2}{2 a}\right] \\
& =\frac{1}{2 \mathrm{R}}\left[\frac{a^2+b^2-c^2-a^2-c^2+b^2}{2 a}\right]=\frac{1}{2 \mathrm{R}}\left(\frac{2 b^2-2 c^2}{2 a}\right) \\
& =\frac{1}{2 R}\left(\frac{b^2-c^2}{a}\right) \\
& \therefore \quad \mathrm{RHS}=\frac{\mathrm{Nr}}{\mathrm{Dr}}=\frac{1}{2 \mathrm{R}}\left(\frac{a^2-b^2}{c}\right) / \frac{1}{2 \mathrm{R}}\left(\frac{b^2-c^2}{a}\right) \\
& =\frac{a}{c}\left(\frac{a^2-b^2}{b^2-c^2}\right) \\
& \text { Given LHS }=\text { RHS } \Rightarrow \frac{a}{c}=\frac{a}{c}\left(\frac{a^2-b^2}{b^2-c^2}\right) \Rightarrow \frac{a^2-b^2}{h^2-r^2}=1 \\
& a^2-b^2=b^2-c^2 \Rightarrow a^2+c^2=2 b^2 \\
& \Rightarrow a^2, b^2, c^2 \text { are in A.P. } \\
&
\end{aligned}
$
Question 2.
The angles of a triangle $\mathrm{ABC}$, are in Arithmetic Progression and if $\mathrm{b}: \mathrm{c}=\sqrt{3}: \sqrt{2}$, find $\angle \mathrm{A}$.

Solution:
Given $\frac{b}{c}=\frac{\sqrt{3}}{\sqrt{2}}$ and $\angle \mathrm{A}, \angle \mathrm{B}$ and $\angle \mathrm{C}$ are in A.P.
$
\begin{array}{ll}
\therefore & 2 \angle \mathrm{B}=\angle \mathrm{A}+\angle \mathrm{C} \\
\Rightarrow & 3 \angle \mathrm{B}=\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}
\end{array}
$
Now from sine formula, we have
$
\begin{aligned}
\frac{b}{\sin \mathrm{B}} & =\frac{c}{\sin \mathrm{C}} \\
\frac{b}{c} & =\frac{\sin \mathrm{B}}{\sin \mathrm{C}} \Rightarrow \frac{\sqrt{3}}{\sqrt{2}}=\frac{\sin 60^{\circ}}{\sin \mathrm{C}} \\
\Rightarrow \quad & \frac{\sqrt{3}}{\sqrt{2}}=\frac{\frac{\sqrt{3}}{2}}{\sin \mathrm{C}}
\end{aligned}
$

Question 3.
In a $\triangle A B C$, if $\cos C=\frac{\sin A}{2 \sin B}$, show that the triangle is isosceles.
Solution:
Let $\frac{\sin \mathrm{A}}{a}=\frac{\sin \mathrm{B}}{b}=\frac{\sin \mathrm{C}}{c}=k$ (say)
$
\begin{gathered}
\Rightarrow \quad \sin \mathrm{A}=a k, \sin \mathrm{B}=b k \text { and } \sin \mathrm{C}=c k \text { we are given } \cos \mathrm{C}=\frac{\sin \mathrm{A}}{2 \sin \mathrm{B}} \\
\Rightarrow \quad \frac{a^2+b^2-c^2}{2 a b}=\frac{a k}{2 b k}=\frac{a}{2 b} \\
\Rightarrow \quad \frac{a^2+b^2-c^2}{2 a}=\frac{a}{2} \\
\Rightarrow \quad \frac{a^2+b^2-c^2}{a}=\frac{a}{1} \\
\Rightarrow \mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2=\mathrm{a}^2 \Rightarrow \mathrm{b}^2-\mathrm{c}^2=\mathrm{a}^2-\mathrm{a}^2 \\
\Rightarrow \mathrm{b}^2-\mathrm{c}^2=0 \Rightarrow \mathrm{b}=\mathrm{c}
\end{gathered}
$
$\therefore \triangle \mathrm{ABC}$ is isosceles

Question 4.
In a $\triangle A B C$, prove that $\frac{\sin B}{\sin C}=\frac{c-a \cos B}{b-a \cos C}$.
Solution:
$
\begin{aligned}
& \mathrm{RHS}=\frac{c-a \cos \mathrm{B}}{b-a \cos \mathrm{C}} \\
& \text { Nr. } c-a \cos \mathrm{B}=c-a\left(\frac{a^2+c^2-b^2}{2 a c}\right) \\
& =c-\left(\frac{a^2+c^2-b^2}{2 c}\right)=\frac{2 c^2-a^2-c^2+b^2}{2 c} \\
& =\frac{b^2+c^2-a^2}{2 c}=\frac{b^2+c^2-a^2}{2 c} \times \frac{b}{b} \\
& =b\left\{\frac{b^2+c^2-a^2}{2 b c}\right\}=b \cos \mathrm{A} \\
& \text { Dr. } b-a \cos \mathrm{C}=b-a\left[\frac{a^2+b^2-c^2}{2 a b}\right] \\
& =b-\left(\frac{a^2+b^2-c^2}{2 b}\right)=\frac{2 b^2-a^2-b^2+c^2}{2 b} \\
& =\frac{b^2+c^2-a^2}{2 b}=\frac{b^2+c^2-a^2}{2 b} \times \frac{c}{c} \\
& =c\left[\frac{b^2+c^2-a^2}{2 b c}\right]=c \cos \mathrm{A} \\
& \therefore \mathrm{RHS}=\frac{b \cos \mathrm{A}}{c \cos \mathrm{A}}=\frac{b}{c} \\
& =\frac{2 \mathrm{R} \sin \mathrm{B}}{2 \mathrm{R} \sin \mathrm{C}}=\frac{\sin \mathrm{B}}{\sin \mathrm{C}}=\mathrm{LHS} \\
&
\end{aligned}
$

Question 5 .
In a $\triangle \mathrm{ABC}$, prove that a $\cos \mathrm{A}+\mathrm{b} \cos \mathrm{B}+\mathrm{c} \cos \mathrm{C}=2 \mathrm{a} \sin \mathrm{B} \sin \mathrm{C}$.
Solution:
LHS $=\mathrm{a} \cos \mathrm{A}+6 \cos \mathrm{B}+\mathrm{c} \cos \mathrm{C}$
Using sine formula, we get $\mathrm{k} \sin \mathrm{A} \cos \mathrm{A}+\mathrm{k} \sin \mathrm{B} \cos \mathrm{B}+\mathrm{k} \sin \mathrm{C} \cos \mathrm{C} k$
$=\frac{k}{2}[2 \sin \mathrm{A} \cos \mathrm{A}+2 \sin \mathrm{B} \cos \mathrm{B}+2 \sin \mathrm{C} \cos \mathrm{C}]$
$=\frac{k}{2}[\sin 2 \mathrm{~A}+\sin 2 \mathrm{~B}+\sin 2 \mathrm{C}]$
$=\frac{k}{2}[2 \sin (\mathrm{A}+\mathrm{B}) \cdot \cos (\mathrm{A}-\mathrm{B})+2 \sin \mathrm{C} \cdot \cos \mathrm{C}]$
$=\frac{k}{2}[2 \sin (\mathrm{A}-\mathrm{B}) \cdot \cos (\mathrm{A}-\mathrm{B})+2 \sin \mathrm{C} \cdot \cos \mathrm{C}]$
$=\frac{k}{2}[2 \sin \mathrm{C} \cdot \cos (\mathrm{A}-\mathrm{B})+2 \sin \mathrm{C} \cdot \cos \mathrm{C}]$
$=\mathrm{k} \sin \mathrm{C}[\cos (\mathrm{A}-\mathrm{B})+\cos \mathrm{C}]$
$=k \sin \mathrm{C}[\cos (\mathrm{A}-\mathrm{B})+\cos (\pi-\overline{\mathrm{A}+\mathrm{B}})]$
$=\mathrm{k} \sin \mathrm{C}[\cos (\mathrm{A}-\mathrm{B})-\cos (\mathrm{A}+\mathrm{B})]$
$=\mathrm{k} \sin \mathrm{C} \cdot 2 \sin \mathrm{A} \sin \mathrm{B}$
$=2 \mathrm{k} \sin \mathrm{A} \cdot \sin \mathrm{B} \sin \mathrm{C}$
$=2 \mathrm{a} \sin \mathrm{B} \sin \mathrm{C}=\mathrm{RHS}$

Question 6.
In a $\triangle \mathrm{ABC}, \angle \mathrm{A}=60^{\circ}$. Prove that $b+c=2 a \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)$
Solution:
$
\begin{aligned}
\mathrm{LHS} & =b+c=2 \mathrm{R} \sin \mathrm{B}+2 \mathrm{R} \sin \mathrm{C} \\
& =2 \mathrm{R}[\sin \mathrm{B}+\sin \mathrm{C}]=2 \mathrm{R}\left[2 \sin \frac{\mathrm{B}+\mathrm{C}}{2} \cos \frac{\mathrm{B}-\mathrm{C}}{2}\right] \\
& =4 \mathrm{R} \cos \frac{\mathrm{A}}{2} \cos \frac{\mathrm{B}-\mathrm{C}}{2}=4 \mathrm{R} \frac{\sqrt{3}}{2} \cos \frac{\mathrm{B}-\mathrm{C}}{2} \\
\mathrm{RHS} & =2 a \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2},\right. \\
& =2 \times 2 \mathrm{R} \sin \mathrm{A} \cos \frac{\mathrm{B}-\mathrm{C}}{2} \\
& =4 \mathrm{R} \sin 60^{\circ} \cos \frac{\mathrm{B}-\mathrm{C}}{2} \\
& =4 \mathrm{R}\left(\frac{\sqrt{3}}{2}\right) \cos \frac{\mathrm{B}-\mathrm{C}}{2} \\
(1)=(2) & \Rightarrow \mathrm{LHS}=\mathrm{RHS}
\end{aligned}
$

Question 7.
In a $\triangle \mathrm{ABC}$, prove the following.
(i) $a \sin \left(\frac{A}{2}+B\right)=(b+c) \sin \frac{A}{2}$
(ii) $a(\cos \mathrm{B}+\cos \mathrm{C})=2(b+c) \sin ^2 \frac{\mathrm{A}}{2}$
(iii) $\frac{a^2-c^2}{b^2}=\frac{\sin (\mathrm{A}-\mathrm{C})}{\sin (\mathrm{A}+\mathrm{C})}$
(iv) $\frac{a \sin (\mathrm{B}-\mathrm{C})}{b^2-c^2}=\frac{b \sin (\mathrm{C}-\mathrm{A})}{c^2-a^2}=\frac{c \sin (\mathrm{A}-\mathrm{B})}{a^2-b^2}$
(v) $\frac{a+b}{a-b}=\tan \left(\frac{A+B}{2}\right) \cot \left(\frac{A-B}{2}\right)$

Solution:
(i)
$
\begin{aligned}
& \frac{a}{b+c}=\frac{2 \mathrm{R} \sin \mathrm{A}}{2 \mathrm{R} \sin \mathrm{B}+2 \mathrm{R} \sin \mathrm{C}} \\
& =\frac{2 R(\sin A)}{2 R(\sin B+\sin C)}=\frac{\sin A}{\sin B+\sin C} \\
& =\frac{2 \sin \mathrm{A} / 2 \cos \mathrm{A} / 2}{2 \sin \frac{\mathrm{C}+\mathrm{B}}{2} \cos \frac{\mathrm{C}-\mathrm{B}}{2}}=\frac{\sin \mathrm{A} / 2 \cos \mathrm{A} / 2}{\cos \mathrm{A} / 2 \cos \frac{\mathrm{C}-\mathrm{B}}{2}} \\
& =\frac{\sin \mathrm{A} / 2}{\sin 90^{\circ}-\left(\frac{\mathrm{C}-\mathrm{B}}{2}\right)}=\frac{\sin \mathrm{A} / 2}{\sin \frac{\mathrm{A}+\mathrm{B}+\mathrm{C}}{2}-\frac{(\mathrm{C}-\mathrm{B})}{2}} \\
& =\frac{\sin \mathrm{A} / 2}{\sin \left(\frac{\mathrm{A}}{2}+\frac{\mathrm{B}}{2}+\frac{\mathrm{C}}{2}-\frac{\mathrm{C}}{2}+\frac{\mathrm{B}}{2}\right)}=\frac{\sin \mathrm{A} / 2}{\sin \left(\frac{\mathrm{A}}{2}+\mathrm{B}\right)} \\
& \frac{a}{b+c}=\frac{\sin \mathrm{A} / 2}{\sin \left(\frac{\mathrm{A}}{2}+\mathrm{B}\right)} \\
& \therefore a \sin \left(\frac{\mathrm{A}}{2}+\mathrm{B}\right)=(b+c) \sin \frac{\mathrm{A}}{2} \\
&
\end{aligned}
$

(ii)
$
\begin{aligned}
\frac{a}{b+c} & =\frac{2 \mathrm{R} \sin \mathrm{A}}{2 \mathrm{R} \sin \mathrm{B}+2 \mathrm{R} \sin \mathrm{C}}=\frac{\sin \mathrm{A} / 2}{\cos \frac{\mathrm{B}-\mathrm{C}}{2}} \mathrm{fr} \\
\text { Now } \frac{2 \sin ^2 \mathrm{~A} / 2}{\cos \mathrm{B}+\cos \mathrm{C}} & =\frac{2 \sin ^2 \mathrm{~A} / 2}{2 \cos \frac{\mathrm{B}+\mathrm{C}}{2} \cos \frac{\mathrm{B}-\mathrm{C}}{2}} \\
& =\frac{2 \sin ^2 \frac{\mathrm{A}}{2}}{2 \sin \frac{\mathrm{A}}{2} \cos \frac{\mathrm{B}-\mathrm{C}}{2}}=\frac{\sin \frac{\mathrm{A}}{2}}{\cos \frac{\mathrm{B}-\mathrm{C}}{2}} \\
(1)=(2) \Rightarrow \frac{a}{b+c} & =\frac{2 \sin \frac{\mathrm{A}}{2}}{\cos \mathrm{B}+\cos \mathrm{C}} \\
\Rightarrow a(\cos \mathrm{B}+\cos \mathrm{C}) & =2(b+c) \sin ^2 \frac{\mathrm{A}}{2}
\end{aligned}
$

(iii) Using sine formula
$
\begin{aligned}
\text { LHS }=\frac{a^2-c^2}{b^2} & =\frac{k^2 \sin ^2 \mathrm{~A}-k^2 \sin ^2 \mathrm{C}}{k^2 \sin ^2 \mathrm{~B}} \\
& =\frac{\sin ^2 \mathrm{~A}-\sin ^2 \mathrm{C}}{\sin ^2 \mathrm{~B}}=\frac{\sin (\mathrm{A}+\mathrm{C}) \cdot \sin (\mathrm{A}-\mathrm{C})}{\sin ^2 \mathrm{~B}} \\
& =\frac{\sin \mathrm{B} \sin (\mathrm{A}-\mathrm{C})}{\sin ^2 \mathrm{~B}}=\frac{\sin (\mathrm{A}-\mathrm{C})}{\sin \mathrm{B}} \\
& =\sin (\mathrm{A}-\mathrm{C}) / \sin \left(180^{\circ}-\mathrm{A}+\mathrm{C}\right) \\
& =\frac{\sin (\mathrm{A}-\mathrm{C})}{\sin (\mathrm{A}+\mathrm{C})}=\mathrm{RHS}
\end{aligned}
$

(iv)

$
\begin{aligned}
& a \sin (\mathrm{B}-\mathrm{C})=a[\sin \mathrm{B} \cos \mathrm{C}-\cos \mathrm{B} \sin \mathrm{C}] \\
& =a\left[\frac{b}{2 \mathrm{R}}\left(\frac{a^2+b^2-c^2}{2 a b}\right)-\frac{a^2+c^2-b^2}{2 a c}\left(\frac{c}{2 \mathrm{R}}\right)\right] \\
& =\frac{1}{2 R}\left(\frac{a^2+b^2-c^2}{2}-\frac{a^2+c^2-b^2}{2}\right) \\
& =\frac{1}{4 \mathrm{R}}\left(a^2+b^2-c^2-a^2-c^2+b^2\right) \\
& =\frac{2\left(b^2-c^2\right)}{4 \mathrm{R}} \\
& \therefore \quad \frac{a \sin (\mathrm{B}-\mathrm{C})}{b^2-c^2}=\frac{2\left(b^2-c^2\right)}{4 \mathrm{R}} / b^2-c^2=\frac{2}{4 \mathrm{R}}=\frac{1}{2 \mathrm{R}} \\
& b(\sin \mathrm{C}-\mathrm{A})=b[\sin \mathrm{C} \cos \mathrm{A}-\cos \mathrm{C} \sin \mathrm{A}) \\
& =b\left[\frac{c}{2 \mathrm{R}}\left(\frac{b^2+c^2-a^2}{2 b c}\right)-\frac{a^2+b^2-c^2}{2 a b} \times \frac{a}{2 \mathrm{R}}\right] \\
& =\frac{1}{4 \mathrm{R}}\left[b^2+c^2-a^2-a^2-b^2+c^2\right] \\
& =\frac{1}{4 \mathrm{R}}\left(2\left(c^2-a^2\right)\right)=\frac{c^2-a^2}{2 \mathrm{R}} \\
& \therefore \quad \frac{b \sin (\mathrm{C}-\mathrm{A})}{c^2-a^2}=\frac{c^2-a^2}{2 \mathrm{R}} / c^2-a^2=\frac{1}{2 \mathrm{R}} \\
&
\end{aligned}
$

Similarly $\frac{\sin (\mathrm{A} \quad \mathrm{B})}{a^2 b^2}=\frac{1}{2 \mathrm{R}}$
$
\therefore \frac{a \sin (\mathrm{B}-\mathrm{C})}{b^2-c^2}=\frac{b \sin (\mathrm{C}-\mathrm{A})}{c^2-a^2}=\frac{c \sin (\mathrm{A}-\mathrm{B})}{a^2-b^2}=\frac{1}{2 \mathrm{R}}
$
(v)
$
\begin{aligned}
\mathrm{LHS} & =\frac{a+b}{a-b}=\frac{2 \mathrm{R} \sin \mathrm{A}+2 \mathrm{R} \sin \mathrm{B}}{2 \mathrm{R} \sin \mathrm{A}-2 \mathrm{R} \sin \mathrm{B}} \\
& =\frac{2 \mathrm{R}(\sin \mathrm{A}+\sin \mathrm{B})}{2 \mathrm{R}(\sin \mathrm{A}-\sin \mathrm{B})}=\frac{2 \sin \frac{\mathrm{A}+\mathrm{B}}{2} \cos \frac{\mathrm{A}-\mathrm{B}}{2}}{2 \cos \frac{\mathrm{A}+\mathrm{B}}{2} \sin \frac{\mathrm{A}-\mathrm{B}}{2}} \\
& =\tan \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cot \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=\mathrm{RHS}
\end{aligned}
$
Question 8.
In a $\triangle A B C$, prove that $\left(a^2-b^2+c^2\right) \tan B=\left(a^2+b^2-c^2\right) \tan C$

Solution:
$
\begin{aligned}
{\left[\frac{a^2+c^2-b^2}{2 a c}\right] } & =\cos \mathrm{B} \\
\text { So, } a^2+c^2-b^2 & =2 a c \cos \mathrm{B} \\
\therefore \text { LHS }\left(a^2+c^2-b^2\right) \tan \mathrm{B} & =2 a c \cos \mathrm{B} \times \tan \mathrm{B} \\
& =2 a c \cos \mathrm{B} \times \frac{\sin \mathrm{B}}{\cos \mathrm{B}} \\
& =2 a c \sin \mathrm{B}=2 a c \times \frac{b}{2 \mathrm{R}} \\
& =\frac{2 a b c}{2 \mathrm{R}}=\frac{a b c}{\mathrm{R}} \\
\frac{a^2+b^2-c^2}{2 a b} & =\cos \mathrm{C} \\
\therefore a^2+b^2-c^2 & =2 a b \cos \mathrm{C} \\
\text { RHS }=\left(a^2-c^2+b^2\right) \tan \mathrm{C} & =2 a b \cos \mathrm{C} \tan \mathrm{C} \\
& =2 a b \cos \mathrm{C} \frac{\sin \mathrm{C}}{\cos \mathrm{C}}=2 a b \sin \mathrm{C} \\
& =2 a b \times \frac{c}{2 \mathrm{R}}=\frac{a b c}{\mathrm{R}}
\end{aligned}
$
$
(1)=(2) \Rightarrow \text { LHS }=\text { RHS }
$
Question 9
An Engineer has to develop a triangular shaped park with a perimeter $120 \mathrm{~m}$ in a village. The park to be developed must be of maximum area. Find out the dimensions of the park.
Solution:
Given, the perimeter of triangular shaped park $=120 \mathrm{~m}$

Perimeter of triangle $=a+b+c=120 \mathrm{~m}$ and we know, $\frac{a+b+c}{2}=\mathrm{s}$ $\therefore 2 \mathrm{~s}=120 \mathrm{~m} \Rightarrow \mathrm{s}=60 \mathrm{~m}$
Maximum area of triangular shaped park $\leq \frac{s^3}{27}$

$
\therefore(s-a)(s-b)(s-c) \leq \frac{s^3}{27}
$
When area is maximum, then $a=b=c$
$
\begin{aligned}
& \therefore(s-a)^3 \leq \frac{s^3}{27} \\
& \therefore(s-a)^3 \leq\left(\frac{s}{3}\right)^3 \\
& \therefore s-a \leq \frac{s}{3} \\
& 60-a \leq \frac{60}{3}(\text { substitute } s=600) \\
& 60-a \leq 20 \quad \therefore 40 \leq a
\end{aligned}
$
All sides of a triangular part would be $40 \mathrm{~m}$.
i.e., $a=40 \mathrm{~m}$,
$\mathrm{b}=40 \mathrm{~m}$,
$
\mathrm{c}=40 \mathrm{~m} \text {. }
$

Question 10 .
A rope of length $12 \mathrm{~m}$ is given. Find the largest area of the triangle formed by this rope and find the dimensions of the triangle so formed.
Solution:
The largest triangle will be an equilateral triangle
$
\begin{aligned}
& \therefore \text { Side of the triangle }=\frac{12}{3}=4 \mathrm{~m}=a \\
& \text { Area of the triangle }=\frac{a^2 \sqrt{3}}{4}=\frac{4^2 \sqrt{3}}{4}=4 \sqrt{3} \mathrm{sq} \cdot \mathrm{m}
\end{aligned}
$
Question 11.
Derive Projection formula from
(i) Law of sines,
(ii) Law of cosines.
Solution:
(i) To Prove $a=b \cos \mathrm{c}+\mathrm{c} \cos \mathrm{B}$
Using sine formula
$\mathrm{RHS}=\mathrm{b} \cos \mathrm{C}+\mathrm{c} \cos \mathrm{B}$
$=2 \mathrm{R} \sin \mathrm{B} \cos \mathrm{C}+2 \mathrm{R} \sin \mathrm{C} \cos \mathrm{B}$
$=2 R[\sin B \cos C+\cos B \sin C]$
$
\begin{aligned}
& =2 R \sin (B+C)=2 R[\sin \pi-\mathrm{A}) \\
& =2 \mathrm{R} \sin \mathrm{A}=\mathrm{a}=\mathrm{LHS}
\end{aligned}
$

(ii) To prove $\mathrm{a}=\mathrm{b} \cos \mathrm{c}+\mathrm{c} \cos \mathrm{B}$
Using cosine formula
$
\begin{aligned}
\text { RHS } & =b \cos \mathrm{C}+c \cos \mathrm{B} \\
& =b\left[\frac{a^2+b^2-c^2}{2 a b}\right]+c\left[\frac{a^2+c^2-b^2}{2 a c}\right] \\
& =\frac{a^2+b^2-c^2}{2 a}+\frac{a^2+c^2-b^2}{2 a} \\
& =\frac{a^2+b^2-c^2+a^2+c^2-b^2}{2 a}=\frac{2 a^2}{2 a} \\
& =a=\text { LHS }
\end{aligned}
$