>> TsU = 40° and QS = QU and RS = RT
<< what the ∠RpQ ?
Solution
we'll use lower case for corner angles
Isosceles triangle's base angles are equal
full triandle angle is 180°
RTS:
(1) r + RtS + RsT = 180°
(2) RtS = RsT
QUS:
(3) q + QuS + QsU = 180°
(4) QuS = QsU
PRQ:
(5) p + r + q = 180°
Angle on a straight line is 180°:
(6) RsT + (TsU = 40°) + QsU = 180°
Implies:
(7) 2 RsT = 180° - r // from (1)
(8) 2 QsU = 180° - q // from (3)
(6') RsT + QsU = 140°
Solve the system and find the r + q:
(180° - r)/2 + (180° - q)/2 = 140° =>
=> - (r + q) = -(2 * 180°) + 2 * 140° =>
=> r + q = 2 (180° - 140°) =>
=> r + q = 80°
Using (5) we have the ∠p:
80° + p = 180° =>
p = 100°
p = 100°
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