P, Q & R form a right-angled triangle. R, S & P lie on a straight line.
PS = SQ and ∠SQR = 36°.
Work out
∠RPQ

∠RPQ = 27

Step-by-step explanation:

In ΔSRQ,

∠R = 90

∠SQR = 36°

∠R + ∠SQR + ∠RSQ = 180    {Angle sum property of triangle}

90 + 36 + ∠RSQ = 180

126 + ∠RSQ = 180

         ∠RSQ = 180 - 126

        ∠RSQ = 54°

∠PSQ +∠RSQ = 180   {Linear pair}

∠PSQ + 54 = 180

 ∠PSQ = 180 - 54

∠PSQ = 126

In ΔPSQ,

SQ = PS ,

So, ∠SQP = ∠SPQ    {Angles opposite to equal sides are equal}

∠SQP = ∠SPQ   =x

∠PSQ + x +x = 180  {Angle sum property of triangle}

126 + 2x = 180

2x = 180 - 126

2x = 54

x = 54/2

x = 27

∠RPQ = 27°

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !