Question

In the given figure, PQ and RS intersect at point O, PQ and TU intersect at M and TU and RS intersect at point N. If ∠POR : ∠ROM = 1 : 5 and ∠UMQ : ∠UMO = 2 : 7, then the sum of x, y and z is

• A. 380°
• B. 390°
• C. 360°
• D. 400°

Answer 1

The correct option is D 400°

Given: ∠POR : ∠ROM = 1 : 5 and ∠UMQ : ∠UMO = 2 : 7
$\angle POR+\angle ROM={180}^{\circ }$ (Linear pair)
$\Rightarrow \frac{1}{5}\angle ROM+\angle ROM={180}^{\circ }$ $(\because \angle POR:\angle ROM=1:5)$
$\Rightarrow \frac{6}{5}\angle ROM={180}^{\circ }$
$\Rightarrow \angle ROM=\frac{5}{6}\times {180}^{\circ }$
$\Rightarrow \angle ROM={150}^{\circ }$
Also, ∠PON = ∠ROM (Vertically opposite angles)
$\Rightarrow x={150}^{\circ }$ …..(i)
Now, $\angle UMQ+\angle UMO={180}^{\circ }$ (Linear pair)
$\Rightarrow \frac{2}{7}\angle UMO+\angle UMO={180}^{\circ }$ $(\because \angle UMQ:\angle UMO=2:7)$
$\Rightarrow \frac{9}{7}\angle UMO={180}^{\circ }$
$\Rightarrow \angle UMO=\frac{7}{9}\times {180}^{\circ }$
$\Rightarrow \angle UMO={140}^{\circ }$
Also, $\angle QMN=\angle UMO$ (Vertically opposite angles)
$\Rightarrow y={140}^{\circ }$ …..(ii)
Now, $\angle MON+\angle PON={180}^{\circ }$ (Linear pair)
$\Rightarrow \angle MON={180}^{\circ }–{150}^{\circ }$ (From (i)]
$\Rightarrow \angle MON={30}^{\circ }$
and $\angle ONM=\angle TNS$ (Vertically opposite angles)
⇒ $\angle ONM=z$ (∵ $\angle TNS=z$)
In $\Delta OMN$, $\angle MON+\angle ONM=\angle NMQ$ (Exterior angle property)
$\Rightarrow {30}^{\circ }+z=y$
$\Rightarrow z={140}^{\circ }–{30}^{\circ }={110}^{\circ }$ [From (ii)] …..(iii)
$\therefore x+y+z={150}^{\circ }+{140}^{\circ }+{110}^{\circ }$
$={400}^{\circ }$
Hence, the correct answer is option (d).