Q. 44.2( 24 Votes )

# In the adjoining figure, ABCD is a quadrilateral such that ∠D + ∠C = 100°. The bisectors of ∠A and ∠B meet at ∠P. Determine ∠APB.

Answer :

We know that the sum of the angles in a quadrilateral is **360 ^{0}**.

i.e, **∠****A +** **∠****B +** **∠****C +** **∠****D = 360 ^{0}**. ...... (1)

We also know that the sum of angles in a triangle is **180 ^{0}**.

According to the problem, it is given that **∠****C +** **∠****D = 100 ^{0}**.

Substituting the condition in the eq(1) we get,

⇒ ∠A + ∠B + 100^{0} = 360^{0}

⇒ ∠A + ∠B = 360^{0} - 100^{0}

⇒ **∠****A +** **∠****B = 260 ^{0}**.

⇒

⇒ . ...... (2)

According to the problem, it is given that the **ΔABP** is formed by the intersection of angular bisectors of **∠****A** and **∠****B**.

From **ΔABP**, We can write that,

⇒

⇒ 130^{0} + ∠P = 180^{0} (From eq(2))

⇒ ∠P = 180^{0} - 130^{0}

⇒ **∠** **P = 50 ^{0}**.

The value of the ∠P is **50 ^{0}**.

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In the adjoining figure, ABCD is a quadrilateral such that ∠D + ∠C = 100°. The bisectors of ∠A and ∠B meet at ∠P. Determine ∠APB.

Karnataka Board - Mathematics Part II

Two angles of a quadrilateral are 70° and 130° and the other two angles are equal. Find the measure of these two angles.

Karnataka Board - Mathematics Part II