Question

One angle of a triangle is ${61}^0$ and the other two angles are in the ratio of $1\frac{1}{2}:1\frac{1}{3}$. Find these angles.

Answer 1

From the triangle ABC

Consider $\angle A={61}^{\circ }$

In a triangle

$\angle A+\angle \angle B+\angle C={180}^{\circ }$

Substituting the values

${61}^{\circ }+\angle B+\angle C={180}^{\circ }$

By further calculation

$\angle B+\angle C={180}^{\circ }–{61}^{\circ }={119}^{\circ }$

$\angle B:\angle C=1\frac{1}{2}:1\frac{1}{3}=3/2:4/3$

Taking LCM

$\angle B:\angle C=9:8/6$

$\angle B:\angle C=9:8$

Consider $\angle B=9x$ and $\angle C=8x$

Substituting the values

$9x+8x={119}^{\circ }$

$17x={119}^{\circ }$

$x={119}^{\circ }/17=7^{\circ }$

So we get

$\angle B=9x=9\times 7^{\circ }={63}^{\circ }$

$\angle C=8x=8\times 7^{\circ }={56}^{\circ }$