Asked by Amitabha
In trapezium PQRS, the sides PQ and SR are parallel. Angle RSP is 120 . and RS = SP = 1/3PQ. What is the size of angle PQR ?
Answers
Answered by
Reiny
I made a sketch, and let RS = PS = x
then QP = 3x
Draw RC || SP to meet QP at C
Then CP = x and QC = 2x
PCRS is now a ||gram, making angle QCR = 60°
In triangle RCQ, by the cosine law:
QR^2 = (2x)^2 + x^2 - 2(x)(2x)cos60°
= 5x^2 - 4x^2 (1/2)
= 3x^2
QR = √3 x
In the same triangle, by the sine law:
sinQ/x = sin60/(√3 x)
sinQ = xsin60/(√3x)
= sin60/√3
= (√3/2)/√3 = 1/2
Q = 30°
then QP = 3x
Draw RC || SP to meet QP at C
Then CP = x and QC = 2x
PCRS is now a ||gram, making angle QCR = 60°
In triangle RCQ, by the cosine law:
QR^2 = (2x)^2 + x^2 - 2(x)(2x)cos60°
= 5x^2 - 4x^2 (1/2)
= 3x^2
QR = √3 x
In the same triangle, by the sine law:
sinQ/x = sin60/(√3 x)
sinQ = xsin60/(√3x)
= sin60/√3
= (√3/2)/√3 = 1/2
Q = 30°
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