Question

In the adjoining figure 5.9 $\square$PQRS is a quadrilateral. The bisectors of $\angle$P and $\angle$Q meet at point A. $\angle$S = 50$°$ and $\angle$R = 110$°$, find the measure of $\angle$PAQ.

Answer 1

We are given quadrilateral PQRS in which bisectors of $\angle$P and $\angle$Q meet at point A.
Using the angle addition property of a quadrilateral, we can say:
$\angle$P + $\angle$Q + $\angle$R +$\angle$S = 360°
We know:
$\angle$R = 110° and $\angle$S = 50°
Thus, we have:
$\angle$P + $\angle$Q + $\angle$R + $\angle$S = 360°
$\Rightarrow$ $\angle$P + $\angle$Q + 110^o + 50° = 360°
$\Rightarrow$ $\angle$P + $\angle$Q + 160^o = 360°
$\Rightarrow$$\angle$P + $\angle$Q = 360° – 160^o
$\Rightarrow$ $\angle$P + $\angle$Q = 200°
On multiplying both sides by $\frac{1}{2}$, we get:
$\frac{1}{2}$ × [$\angle$P + $\angle$Q] = $\frac{1}{2}$ × 200°


or, $\frac{1}{2}$ × [$\angle$P + $\angle$Q] = 100°

or, $\angle$APQ + $\angle$AQP = 100° …(1)

Now, in ΔAPQ, we have:
$\angle$APQ + $\angle$AQP + $\angle$PAQ = 180° ...[Angle addition property of triangle]
$\Rightarrow$100° + $\angle$PAQ = 180° [From equation (1)]
$\Rightarrow$$\angle$PAQ = 180° – 100°
$\Rightarrow$$\angle$PAQ= 80°