Question

In the given figure $AB$ is diameter of a circle with centre. If $APQ$ and $RBQ$ are straight lines, $\angle A={35}^o$ and $\angle Q={25}^o$, find : (i) $\angle PRB$ (ii)$\angle PBR$ (iii) $\angle BPR$

Answer 1

Let $AB$ be the diameter of the circle with centre $O$.

Let $ADE$ and $CBE$ be the straight lines that meet at $E$ such that $\angle BAD={35}^{\circ }and\angle BED={25}^{\circ }.$

Join $BD$ and $AC$.

$\left(i\right)$

Now, $\angle BDA={90}^{\circ }$ (Angle in the semicircle)

Also, $\angle EDB+\angle BDA={180}^{\circ }$ (Linear pair)

Or $\angle EDB={90}^{\circ }$

Now, $\angle EBD={180}^{\circ }–(\angle EDB+\angle BED)$ (Angle sum property)

$=({180}^{\circ }–({90}^{\circ }+{25}^{\circ })$

$=({180}^{\circ }–{115}^{\circ })={65}^{\circ }$

$\angle DBC=({180}^{\circ }–\angle EBD)=({180}^{\circ }-{65}^{\circ })={115}^{\circ }$ (Linear pair)

$\therefore \angle DBC={115}^{\circ }$

$\left(ii\right)$

Here, $\angle DCB=\angle BAD$ (Angles in the same segment)

We have $\angle BAD={35}^{\circ }$

$\therefore \angle DCB={35}^{\circ }$

$\left(iii\right)$

$\angle BDC={180}^{\circ }–(\angle DBC+\angle DCB)$ (Angle sum property)

$={180}^{\circ }–(\angle DBC+\angle BAD)$

$={180}^{\circ }–({115}^{\circ }+{35}^{\circ })$

$={180}^{\circ }–{150}^{\circ }$

$={30}^{\circ }$

$\therefore \angle BDC={30}^{\circ }$