Question

in the adjoining figure ABCD is a kite. If BCD =52° and ADB = 42° Find the values of x,y, and z​

Answer 1

Answer:x =42°

y = 96°

z = 64°

Solution:

It is given that AB=AD

hence < ABD =x = 42°

[Because angles opposite to equal sides are equal]

in ∆ABD, apply angle sum property of triangle

$y + 42 + 42 = 180 \\ \\ y = 180 - 84 \\ \\ y = 96$
Now again in ∆BCD,again DC= BC [given]

hence < CBD = <CDB = z

[Because angles opposite to equal sides are equal]

in ∆ABD, apply angle sum property of triangle

$52 + z + z = 180 \\ \\ 2z = 180 - 52 \\ \\ 2z = 128 \\ \\ z = \frac{128}{2} \\ \\ z= 64$
Hope it helps you.

Answer 2

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