Question

can anyone help me doing this 3 questions please ​

Answer 1

Answer:

$85^{0}$

Step-by-step explanation:

∠PRQ = ∠RPY = $45^{0}$ (Alternate angles)

Let ∠QPR = $x^{0}$

Therefore,

∠XPQ + ∠YPR + ∠QPR = $180^{0}$

$50^{0} +45^{0} + x^{0} = 180^{0}$

$95^{0}$ + $x^{0}$ = $180^{0}$

$x^{0} = 180^{0} - 95^{0}$

$x^{0} = 85^{0}$

Therefore,

∠QPR = $85^{0}$