Question

Draw a circle with centre O and radius 3.5 cm. Draw two tangents PA and PB from an external point P such that $<APB={45}^0$. What is the value of $<AOB+<APB$.

Answer 1

Steps of construction are as follows:-

1) Draw a circle of radius OQ= 3.5cm with centre at O.

2) Draw an angle of ${90}^0$ from point O.

3) Let ray of angle intersect the circle at R.

4) Now draw ${90}^0$ from point Q

5) Draw ${90}^0$ at point R.

6) These two arcs meet each other, mark the point as P

PQ and PR are the two tangents inclined at angle of ${45}^0$

Now $\angle AOB={90}^0$ and $\angle APB={45}^0$

$=\angle AOB+\angle APB={135}^0$