Question

In the figure, O is the center of the circle. ∠poq=100° and ∠por=110° then ∠qpr equals


A) 210° b)200 c)150° d)75°

Answer 1

Hello Mate!

Since < POQ = 100° and < POR = 110°

Then, < QOR = < POQ + < POR

< QOR = 100° + 110°

< QOR = 210°

Since ext. < QOR = 2 < QPR ( by theorem ).

Also, ext. < QOR = 360° - int < QOR

ext. < QOR = 360° - 210°

ext. < QOR = 150°

So, 150° = 2 < QPR

75° = < QPR

Therefore Option (D) is correct.

Have great future ahead!

Answer 2

______Heyy Buddy ❤_____

____Here's your Answer ______


Given;

POQ = 100°

POR = 110°

To Find, QPR = ?

Find;

Interior QOR = POQ + POR

=> Int. QOR = 100 + 110

=> Int. QOR = 210°.

Now,

Ext. QOR = Full angle - Int. QOR

=> Ext. QOR = 360 - 210

=> Ext. QOR = 150°.

Now,

we know that, Angle substended at the centre is twice the angle substended at the Circumference.

=> Ext. QOR = 2 × QPR

=> 150 = 2 × QPR

=> QPR = 75°

Therefore,

Option D is correct.
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