Answer:
x = 23
y = 7
z = 11
Step-by-step explanation:
Given ΔPRS ≅ ΔCFH
By CPCTC of these triangles,
m∠R = m∠F
(13y - 1)° = 90°
13y = 90 + 1
13y = 91
y = 7
Side PS ≅ Side CH
2x - 7 = 39
2x = 39 + 7
2x = 46
x = 23
∠H ≅ ∠S
m∠H = m∠S
Since, m∠P + m∠R + m∠S = 180°
m∠S = 180° - (m∠P + m∠R)
= 180° - [28° + (13y - 1)]
= (180 - 28) - (13y - 1)
= 152 - (13y - 1)
Now, 152 - (13y - 1) = (6z - 4) [Since, m∠H = m∠S]
152 - (13×7 - 1) = 6z - 4 [Since, y = 7]
152 - 90 = 6z - 4
62 = 6z - 4
6z = 66
z = 11
