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In the diagram, 0 is the centre of the circle. Find the value x.


Question

In the diagram, 0 is the centre of the circle. Find the value x.

Options

A) 34

B) 29

C) 17

D) 14

The correct answer is D.

Explanation:

POQ in a straight line
Hence, < POQ + < QOR = 180\(^{\circ}\)
56\(^{\circ}\) + < QOR = 180\(^{\circ}\)
< QOR = 180\(^{\circ}\)- 56\(^{\circ}\)
= 124\(^{\circ}\)
Now, in \(\bigtriangleup\) QOR OR = OQ = Radius
< ORQ = < OQR = 2x (Base angles of an Isosceles \(\bigtriangleup\))
2x + 124 + 2x = 180\(^{\circ}\)
4x + 124 = 180
4x = 180 - 124
4x = 56
x = \(\frac{56}{4}\)
x = 14\(^{\circ}\)

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Dicussion (1)

  • POQ in a straight line
    Hence, < POQ + < QOR = 180\(^{\circ}\)
    56\(^{\circ}\) + < QOR = 180\(^{\circ}\)
    < QOR = 180\(^{\circ}\)- 56\(^{\circ}\)
    = 124\(^{\circ}\)
    Now, in \(\bigtriangleup\) QOR OR = OQ = Radius
    < ORQ = < OQR = 2x (Base angles of an Isosceles \(\bigtriangleup\))
    2x + 124 + 2x = 180\(^{\circ}\)
    4x + 124 = 180
    4x = 180 - 124
    4x = 56
    x = \(\frac{56}{4}\)
    x = 14\(^{\circ}\)

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