Question

In figure, $\angle POR=a^o$ and $\angle QOR=b^o$. If $a^o-b^o={60}^o$. Find angle $a^o$ and $b^o$.

• A. $\angle a={70}^o,\angle b={110}^o$
• B. $\angle a={90}^o,\angle b={90}^o$
• C. $\angle a={120}^o,\angle b={60}^o$
• D. $\angle a={60}^o,\angle b={120}^o$

Answer 1

Answer: (C)

$\angle a={120}^o,\angle b={60}^o$

$\angle POR+\angle QOR={180}^{\circ }$

As angle formed on line $POQ$ are supplementary

$\Rightarrow a^{\circ }+b^{\circ }={180}^{\circ }...........\left(i\right)$

Given $\Rightarrow a^{\circ }-b^{\circ }={60}^{\circ }...........\left(ii\right)$

Adding $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow 2a^{\circ }={240}^{\circ }\Rightarrow a^{\circ }={120}^{\circ }$

Substituting $a$ in $\left(i\right)$

$\Rightarrow b^{\circ }={60}^{\circ }$