In the given figure, AB is a diameter of the circle such that ∠A = 35° and ∠Q = 25°, find ∠PBR.
Let us first consider the triangle ΔABQ.
It is known that in a triangle the sum of all the interior angles add up to 180°.
So here in our triangle ΔABQ we have,
`angleBAQ + angleAQB + angleABQ ` = 180°
`angleABQ = 180° - angleBAQ - angle AQB`
= 180° - 35° - 25°
`angle ABQ` = 120°
By a property of the circle we know that an angle formed in a semi-circle will be 90°..
In the given circle since ‘AB’ is the diameter of the circle the angle `angleAPB` which is formed in a semi-circle will have to be 90°.
So, we have `angleAPB` = 90°
Now considering the triangle ΔAPB we have,
`angleAPB + angleBAP + angleABP ` = 180°
`angleABP = 180° - angleAPB - angleBAP`
= 180° - 90° - 35 °
`angleABP` = 55°
From the given figure it can be seen that,
`angleABP + anglePBQ = angle ABQ`
`anglePBQ = angleABQ - angleABP`
= 120°- 55°
`anglePBQ` = 65°
Now, we can also say that,
`anglePBQ + anglePBR` = 180°
`anglePBR = 180° - anglePBQ`
=180° - 65 °
`anglePBR ` = 115°
Hence the measure of the angle `anglePBR` is 115°.
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