Question

A circle with centre O has diameter as AB. If $PA=PB$; find the measure of each angle of the $△APB$.

• A. $\angle PAB={35}^o,\angle APB={90}^o,\angle PBA={55}^o$
• B. $\angle PAB={30}^o,\angle APB={90}^o,\angle PBA={60}^o$
• C. $\angle PAB={45}^o,\angle APB={90}^o,\angle PBA={45}^o$
• D. $\angle PAB={60}^o,\angle APB={90}^o,\angle PBA={30}^o$

Answer 1

The correct option is C $\angle PAB={45}^o,\angle APB={90}^o,\angle PBA={45}^o$

Given- $O$ is the centre of a circle with diameter AB.$\angle APB$ is an angle in the semicircle i.e $AB$ subtends $\angle APB$ at the circumference.
To find out $\begin{array}{c}\angle PAB\angle PBA\angle APB\end{array}$
AB subtends $\angle APB$ at the circumference.
$\therefore \angle APB$ is an angle in the semicircle.
So $\angle APB={90}^o$. Now, in $\Delta APB$, we have $PA=PB.$ i.e $\Delta APB$ is isosceles.
$\therefore \angle PAB=\angle PBA⟹\angle PAB+\angle PBA=2\angle PAB\left(or\angle PBA\right)$.
Again, by angle sum property of triangles, we have
$\angle APB+\angle PAB+\angle PBA={180}^o⟹{90}^o+2\angle PAB={180}^o⟹\angle PAB=\angle PBA={45}^o$.
$\therefore \begin{array}{c}\angle PAB={45}^o\angle APB={90}^o\angle PBA={45}^o\end{array}$
Ans- Option C.