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.

Answer 1

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°.