Question

If sides of triangle are in A.P. and the largest angle is double smallest angle then find ratio of sides.

• A. $3:5:6$
• B. $4:5:6$
• C. $2:3:5$
• D. $3:4:5$

Answer 1

Answer: (B)

$4:5:6$

Let a, b, c are the sides in increasing order $2b=a+c$
Let angles are
$A=\theta ,B=\pi -3\theta$ and $C=2\theta$
Now $2\sin B=\sin A+\sin C$
$2\sin 3\theta =\sin \theta +\sin 2\theta$
$2(3-4{\sin }^2\theta )=1+2\cos \theta$
$6-8(1-{\cos }^2\theta )=1+2\cos \theta$
$8{\cos }^2\theta -2\cos \theta -3=0$
$(2\cos \theta +1)(4\cos \theta -3)=0$
$\cos \theta =\frac{3}{4},\cos \theta =-\frac{1}{2}$ (rejected)
the ratio of sides a $:$ b $:$ c
$\sin A:\sin B:\sin C$
$\sin \theta :\sin 3\theta :\sin 2\theta$
$1:3-4{\sin }^2\theta :2\cos \theta$
$1:\frac{5}{4}:\frac{6}{4}=4:5:6$.