Question

solve the above sum step-by-step ​

Answer 1

Required Answer:

The above question says that:

1. There is a circle with centre O.
2. PM and PN are tangents from a common point P.
3. <MPN is given 60°

And we have to find <MQN in the centre.

So, according to tangent theoram of circles:

"Radius from the center of the circle to the point of tangency is perpendicular to the tangent line."

• <QMN = 90°
• <QNP = 90° according to theoram

Now MQNP is a quadrilateral forming with four angles and we have got three of the angles in the quadrilateral. The tangents are inclined at a angle of 60°. We have to find the fourth angle.

According to angle sum property,

➝ <QMP + <QNP + <MPN + <MQN = 360°

➝ 90° + 90° + 60° + <MQN = 360°

➝ 240° + <MQN = 360°

➝ <MQN = 120°

Hence, the required answer is:

$\large{ \boxed{ \sf{ \purple{ \angle MQN = 120 \degree}}}}$

And we are done !!

Answer 2

$\huge \mathbb{\underbrace{\red{Question\:}}}$

In the adjoining figure,Q. is the centre of the circle and PM,PN are tangent segments to the circle. If ∠MPN = 60° , find ∠MQNThe radius of circle is 7cm and PM = 7 cm. Determine the distance QP.

$\huge{\boxed{\sf Answer}}$

In the given figure of the question, Q is the centre of the circle, and PM and PN are tangents from an external common point 'P'.

∠MPN= 60°, to find ∠MQN

In □PMQN, ∠PMQ=∠PNQ=90°

(Radius is ⊥ to tangent at point of contact from the centre)

According to angle sum property,

➭ <QNP + <MPN+ <QMP + <MQN = 360°

➭ 90° + 60° + 90° + <MQN = 360°

➭ 240° + <MQN = 360°

➭ <MQN = 120°

Hence, the answer is 120°