Question

i request to answer

In the figure given below, ABCD is a kite. If ZBCD = 52º and ZADB = 42 ", find
the value of x, y and z.
​

Answer 1

hey mate here is your answer

hope it helps you

Answer 2

Answer:

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

\begin{gathered}y + 42 + 42 = 180 \\ \\ y = 180 - 84 \\ \\ y = 96 \\ \\\end{gathered}y+42+42=180y=180−84y=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

\begin{gathered}52 + z + z = 180 \\ \\ 2z = 180 - 52 \\ \\ 2z = 128 \\ \\ z = \frac{128}{2} \\ \\ z= 64 \\ \\\end{gathered}52+z+z=1802z=180−522z=128z=2128z=64