Trigonometrical Ratios and Identities 1 Question 3
3. If the lengths of the sides of a triangle are in $AP$ and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is (2019 Main, 8 April II)
(a) $3: 4: 5$
(b) $4: 5: 6$
(c) $5: 9: 13$ (d) $5: 6: 7$
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Answer:
Correct Answer: 3. (b)
Solution:
- Let $a, b$ and $c$ be the lengths of sides of a $\triangle A B C$ such that $a<b<c$.
Since, sides are in AP.
$$ \therefore \quad 2 b=a+c $$
Then, $\quad \angle C=2 \theta$ [according to the question]
So, $\angle B=\pi-3 \theta$
On applying sine rule in Eq. (i), we get
$$ 2 \sin B=\sin A+\sin C $$
$\Rightarrow \quad 2 \sin (\pi-3 \theta)=\sin \theta+\sin 2 \theta \quad$ [from Eq. (ii)]
$\Rightarrow 2 \sin 3 \theta=\sin \theta+\sin 2 \theta$
$\Rightarrow 2\left[3 \sin \theta-4 \sin ^{3} \theta\right]=\sin \theta+2 \sin \theta \cos \theta$
$\Rightarrow 6-8 \sin ^{2} \theta=1+2 \cos \theta \quad[\because \sin \theta$ can not be zero $]$
$\Rightarrow 6-8\left(1-\cos ^{2} \theta\right)=1+2 \cos \theta$
$\Rightarrow 8 \cos ^{2} \theta-2 \cos \theta-3=0$
$\Rightarrow(2 \cos \theta+1)(4 \cos \theta-3)=0$
$\Rightarrow \cos \theta=\frac{3}{4}$
$\operatorname{orcos} \theta=-\frac{1}{2}$ (rejected).
Clearly, the ratio of sides is $a: b: c$
$$ \begin{aligned} & =\sin \theta: \sin 3 \theta: \sin 2 \theta \\ & =\sin \theta:\left(3 \sin \theta-4 \sin ^{3} \theta\right): 2 \sin \theta \cos \theta \\ & =1:\left(3-4 \sin ^{2} \theta\right): 2 \cos \theta \\ & =1:\left(4 \cos ^{2} \theta-1\right): 2 \cos \theta \\ & =1: \frac{5}{4}: \frac{6}{4}=4: 5: 6 \end{aligned} $$
