O is the centre of a circle to which PAX and PBY are tangents from point P at points A and B.Q is a point on the circle,such that angle QAX=49°,angle QBY=62°.what is the measure of angle AQB?
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Given: PAX and PBY are tangents from point P at points A and B, QAX=49°,angle QBY=62°.
To find: angle AQB = ?
Solution:
- Now as we can see that AO and OB are perpendicular to the tangent so, it is a square as the two adjacent sides are equal and the three consecutive angles are equal.
- Now ang OBY = 90
ang OBQ + ang QBY = 90
ang OBQ = 90 - 62
ang OBQ = 28
- Now ang OAX = 90
ang OAQ + ang QAX = 90
ang OAQ = 90 - 49
ang OAQ = 41
- Now outer angle of AOB = 360 - 90 = 270
- In quadrilateral, AOBQ,
A + B + Q + O = 360
Q = 360 - (270 + 41 + 28)
Q = 21°
Answer:
Measure of angle AQB = 21°.
Answered by
4
Answer:
69°
Step-by-step explanation:
Tangents PAX and PBY is straight line
so, angle on each A and B point is 180°
Given <QAX = 49° and <QBY = 62°
Clearly we can say
- <OAP + <QAO + <QAX = 180°
90° + <QAO + 49° = 180°
<QAO = 41°
- <OBP + <QBO + <QBY = 180°
90° + <QBO + 62° = 180°
<QBO = 28°
◆ So, <AQB = 41° + 28° = 69°
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