Question

The magnitudes of the angles A, B, and C in a triangle ABC form an arithmetic progression. The smallest side is a quarter of the largest side. Find the tangent of the smallest angle.

Answer 1

Let the angles be $A=x-d,B=x,C=x+d$. Here A is the smallest angle and hence $BC=a$ is the smallest side.

since $,A,B,C$ are angles of a triangle, $A+B+C=180$

$⟹3x=180$

$⟹x={60}^0$

given that the smallest side is one quarter of the largest side. Thus we have, $\frac{a}{c}=\frac{1}{4}$

using sine rule we know that $\frac{\sin A}{\sin C}=\frac{a}{c}=\frac{1}{4}$

$⟹\frac{sin(x-d)}{sin(x+d)}=\frac{1}{4}$

using componendo and dividendo method, we find that,

$\frac{\sin (x+d)+\sin (x-d)}{\sin (x+d)-\sin (x-d)}=\frac{5}{3}$

$⟹\frac{\sin x\cos d}{\cos x\sin d}=\frac{5}{3}$

$⟹\tan {60}^0\cot d=\frac{5}{3}$

$⟹\tan d=\frac{3\sqrt{3}}{5}$

$\tan A=\tan (x-d)=\frac{\tan x-\tan d}{1+\tan x\tan d}=\frac{\sqrt{3}-\frac{3\sqrt{3}}{5}}{1+\sqrt{3}.\frac{3\sqrt{3}}{5}}=\frac{\sqrt{3}}{7}$