In the diagram above, PQT is an isosceles triangle.|PQ| = |QT|, ∠SRQ = 75°, ∠QPT = 25° and PQR is straight line. Find ∠RST
A.
20o
B.
50o
C.
55o
D.
70o
E.
75o
Correct answer is C
< PTQ = 25° (base angles of an isos. triangle)
\(\therefore\) < PQT = 180° - (25° + 25°) = 130° (sum of angles in triangle PQT)
\(\therefore\) < RST = 130° - 75° = 55° (exterior angle = sum of 2 opp. interior angles)