Question

in the given fig if angle OPQ =25° find angle PMQ
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Answer 1

$\large\mathfrak \pink{Solution:-}$

$\underline \mathbb{GIVEN:-}$

∠OPQ = 25°

$\underline \mathbb{TO \: FIND:-}$

∠PMQ.

$\underline \mathbb{FORMULA:-}$

★Angle on the centre is twice the angle on the circle.

★Angle opposite to equal sides are equal.

$\underline \mathbb{BY \: THE \: PROBLEM:-}$

★OP = OQ( Radius of the same circle.)

So, ∠OPQ = ∠OQP (Angle opposite to equal sides are equal.)

So, ∠OQP = 25°

∠POQ + ∠OPQ + ∠OQP = 180° (Angle sum property of a triangle)

∠POQ + 25° + 25° = 180°

∠POQ = 180° - 25° - 25°

∠POQ = 130°

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∠POQ = 2∠PMQ ( Angle on the centre is twice the angle on the circle. )

130° = 2 ∠ PMQ

So, ∠PMQ = $\cancel\frac{130}{2}$ = 65°

$\underline \mathbb{ANSWER:-}$

∠PMQ = 65°