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Question
In a quadrilateral ABCD, bisectors of angles A and B intersect at O such that ∠AOB = 75°, then write the value of ∠C + ∠D.
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Solution
The quadrilateral can be drawn as follows:
We have AO and BO as the bisectors of angles ∠A and ∠B respectively.
In ΔAOB,We have,
∠AOB + ∠1 +∠2 = 180°
∠AOB = 180° -(∠1 +∠2 )
∠AOB = 180° -`(1/2∠A + 1/2 ∠B)`
∠AOB = 180° - `1/2(∠A +∠B)` …… (1)
By angle sum property of a quadrilateral, we have:
∠A + ∠B +∠C + ∠D = 360°
∠A+ ∠B = 360° -(∠C +∠D)
Putting in equation (1):
`∠AOB =180° - 1/2[360° - (∠C + ∠D )]`
`∠AOB = 180° - 180° +(∠C +∠D)/2]`
`∠AOB = 1/2 (∠C +∠D)` ……(2)
It is given that ∠AOB = 75° in equation (2), we get:
`75° = 1/2 (∠C +∠D)`
`1/2(∠C+∠D) = 75°`
∠C + ∠D = 150°
Hence, the sum of ∠Cand ∠D is 150°.
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