In the figure above, PQR is a straight line segment, PQ = QT. Triangle PQT is an isosceles triangle, ∠SQR is 75o and ∠QPT is 25o. Calculate the value of ∠RST.
A.
45°
B.
55°
C.
25°
D.
55°
Correct Answer: 55°
Explanation
n Δ PQT,
∠PTQ = 25o(base ∠s of isosceles Δ)
In Δ QSR,
∠RQS = ∠QPT + ∠QTP
(Extr = sum of interior opposite ∠s)
∠RQS = 25 + 25
= 50°
Also in Δ QSR,
75 + ∠RQS + ∠QSR = 180o
(sum of ∠s of Δ)
∴75 + 50 + ∠QSR = 180
125 + ∠QSR = 180
∠QSR = 180 - 125
∠QSR = 55°
But ∠QSR and ∠RST are the same
∠RST = 55°
n Δ PQT,
∠PTQ = 25o(base ∠s of isosceles Δ)
In Δ QSR,
∠RQS = ∠QPT + ∠QTP
(Extr = sum of interior opposite ∠s)
∠RQS = 25 + 25
= 50°
Also in Δ QSR,
75 + ∠RQS + ∠QSR = 180o
(sum of ∠s of Δ)
∴75 + 50 + ∠QSR = 180
125 + ∠QSR = 180
∠QSR = 180 - 125
∠QSR = 55°
But ∠QSR and ∠RST are the same
∠RST = 55°
