Question

The circular measures of two angles of a triangle are $\frac{1}{2}$ and $\frac{1}{3}$, find the third angle in English system.

• A. $130^{\circ }15^{\prime }20^{\prime \prime }$
• B. $132^{\circ }15^{\prime }20^{\prime \prime }$
• C. $132^{\circ }16^{\prime }22^{\prime \prime }$
• D. $122^{\circ }16^{\prime }44^{\prime \prime }$

Answer 1

Answer: (C)

We know that the sum of three angles of a triangle is $180^{\circ }$, i.e., $\pi$ radians
$\therefore$ The third angle $=\left(\pi -\frac{1}{2}-\frac{1}{3}\right)$ radian
$\simeq {\left(\frac{22}{7}-\frac{1}{2}-\frac{1}{3}\right)}^c={\left(\frac{97}{42}\times \frac{180}{\pi }\right)}^{\circ }\left(\because {\pi }^c=180^{\circ }\right)$
$={\left(\frac{97\times 30}{22}\right)}^{\circ }=\frac{1455}{11}$ degree
$={\left(132\frac{3}{11}\right)}^{\circ }$
$=132^{\circ }{\left(\frac{3\times 60}{11}\right)}^{\prime }$ $\left(\because 1^{\circ }=60^{\prime }\right)$
$=132^{\circ }{\left(16+\frac{4}{11}\right)}^{\prime }$
$=132^{\circ }16^{\prime }{\left(\frac{4}{11}\times 60\right)}^{\prime \prime }\left(\because 1^{\prime }=60^{\prime }\right)$
$=132^{\circ }16^{\prime }22^{\prime \prime }$.