Question

If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is :

• A. $5:6:7$
• B. $5:9:13$
• C. $4:5:6$
• D. $3:4:5$

Answer 1

The correct option is C $4:5:6$

Let the smallest angle be $A$
and let the sides in ascending order be $a,b,c$
Then the corresponding angles are $A,{180}^{\circ }-3A,2A$

$a,b,c$ are in A.P.
$\therefore 2b=a+c$
$\Rightarrow 2\times 2R\sin ({180}^{\circ }-3A)=2R\sin A+2R\sin 2A$
$\Rightarrow 2\sin 3A=\sin A+\sin 2A\cdots \left(1\right)$
$\Rightarrow 6\sin A-8{\sin }^{3}A=\sin A+2\sin A\cos A$
$\Rightarrow 5\sin A-2\sin A\cos A-8{\sin }^{3}A=0$
$\Rightarrow \sin A(-8{\sin }^{2}A-2\cos A+5)=0$
$\Rightarrow 8{\cos }^{2}A-2\cos A-3=0$
$\Rightarrow \cos A=\frac{3}{4}$
or $\cos A=\frac{-1}{2}\left(\text{not possible}\right)$
$\therefore \sin A=\frac{\sqrt{7}}{4}$

$\sin 2A=2\sin A\cos A=\frac{3\sqrt{7}}{8}$

$\sin 3A=\frac{\sin A+\sin 2A}{2}=\frac{5\sqrt{7}}{16}$
$\therefore a:b:c=\sin A:\sin 3A:\sin 2A$
$=\frac{\sqrt{7}}{4}:\frac{5\sqrt{7}}{16}:\frac{3\sqrt{7}}{8}$
$=4:5:6$