Question

perpendicular to each other.
In figure, PQ is the diameter of the circle with centre O. If ZPQR=65°, ZRPS = 40° and ZPQM=50°, find
ZQPR, ZPRS and ZQPM.
S.
R
40°
Р
65%
Q
50°
M
no
Xand AD produced​

Answer 1

Answer:

∠QMP = ∠QRP = ∠PSR =  90 (Angle subtended by a semi circle is 90 degrees)

In triangle PQM

∠QMP + ∠MQP + ∠QPM = 180 (Angle sum property of traingle)

90 + 50 + ∠QPM = 180

140 + ∠QPM = 180

∠QPM = 180 - 140

∠QPM = 40

In triangle PQR

∠PQR + ∠QRP + ∠QPR = 180 (Angle sum property of traingle)

65 + 90 + ∠QPR = 180

155 + ∠QPR = 180

∠QPR = 25

In triangle PRS

∠PRS + ∠PSR + ∠RPS = 180 (Angle sum propery of triangle)

∠PRS + 90 + 40 = 180

∠PRS + 130 = 180

∠PRS = 180 - 130

∠PRS = 50

Hope this helps you :)

Don't forget to add the degree symbol