Question

How caan we find x and y

Answer 1

$\underline{\mathfrak{\huge{Question:}}}$

Find the values of /_x and /_y in the given.

The Figure has been attached in the answer.

$\underline{\mathfrak{\huge{Answer:}}}$

◇$\tt{Part\:A}$

▪$\sf{Given:}$

AB = CB

/_B = 90°

▪$\sf{Solution:}$

In Triangle ABC :-

/_BCA = /_x ( Opposite angle of equal sides, Given ) ...(1)

=》 /_x + /_BCA + /_B = 180° ( Angle Sum Property )

=》 /_x + /_x + 90° = 180° ( Due to (1) )

=》 2/_x = 90°

=》 $\tt{x = 45}$°

▪/_y = /_B + /_x ( exterior angle property )

=》 /_y = 90° + 45°

=》 $\tt{y = 135}$°

$\tt{Part\:B}$

▪$\sf{Given:}$

PQ = PR

/_SPT = 85°

▪$\sf{Solution:}$

In Triangle PQR :-

/_SPT = /_QPR = 85° ( vertically opposite angles ) ....(2)

/_x = /_PRQ ( Opposite angles of equal sides, Given ) ...(3)

=》 /_x + /_QPR + /_PRQ = 180° ( Angle Sum Property)

=》 /_x + 85° + /_x = 180° ( Due to (2) and (3) )

=》 2/_x = 95°

=》 $\tt{x = 45.5}$°

/_y = /_QPR + /_x ( exterior angle property )

=》 /_y = 85° + 45.5°

=》 $\tt{y = 130.5}$°

That's your answer !