Question

Please solve this problem fast​

Answer 1

Answer:

question 1)

given angle pqt =110degrees,and angle spr=135 degrees

to find- prq

angle QPR+angle SPR=180 degrees

= angle QPR= 180-135=45 degrees

similarly angle PQR=180-TQP

= 180-110=70 degrees

by angle sum property

PQR+QPR+PRQ= 180 degrees

= 180-(70+45)= 65 degrees

Answer 2

Answer:

1) Given-

∠ SPR = 135°

∠ PQT = 110°

To Find- ∠ PRQ

Solution -

∠ RPQ + ∠ SPR = 180° ............linear pair

∠ RPQ + 135 = 180

∠ RPQ = 180 - 135

∠ RPQ = 45°

∠PQT + ∠PQR = 180°...........linear pair

110 + ∠PQR = 180

∠PQR = 180 - 110

∠PQR = 70°

RPQ + PQR + PRQ = 180° .......... angle sum property of triangle

45° + 70° +∠ PRQ = 180°

115 + ∠PRQ = 180

∠PRQ = 180 - 115

∠PRQ = 65°

2)Given-

AD⊥BC

∠BEC = 100°

∠DAC = 45°

To Find - x, y and z

solution-

The place where BE and AD intersect, mark it as O

∠BEC + ∠BEA = 180°...............linear pair

100° + ∠BEA = 180°

∠BEA = 180-100

∠BEA = 80°

In ΔAOE,

∠BEA + ∠DAC + ∠AOE = 180°........angle sum property of triangle

80° + 45° +∠AOE = 180°

125 + ∠AOE  = 180

∠AOE = 180-125

∠AOE = 55°

∠z +∠AOE = 180°.......linear pair

∠z + 55° = 180°

∠z = 180-55

∠z = 125°

In ΔADC,

∠DAC + ∠ADC +∠y = 180°.......angle sum property of triangle

45° + 90° + ∠y = 180°

135 + ∠y = 180

∠y = 180 - 135

∠y = 45°

In ΔBCE,

∠x + ∠y + ∠BEC = 180°.......angle sum property of triangle

∠x + 45° + 100° = 180°

∠x + 145 = 180

∠x = 180-145

∠x = 35°

Hope this helped