Simple geometry problem: angle PQR = θ and angle QPR = 2θ. Prove that cos θ = p/(2q)

Enceladus

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Jan 24, 2023
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Hi - could someone help me with this, please:

In a triangle PQR, angle PQR = θ and angle QPR = 2θ. Prove that cos θ = p/(2q).

This is meant to be solved without using the trig identity sin 2θ = 2 sin θ cos θ, but it is intended for the sine rule to be used. Using the sine rule, I have got as far as

p / sin 2θ = q / sin θ

but I can't get any further. Help much appreciated!
 

stapel

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Feb 4, 2004
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Hi - could someone help me with this, please:

In a triangle PQR, angle PQR = θ and angle QPR = 2θ. Prove that cos θ = p/(2q).

This is meant to be solved without using the trig identity sin 2θ = 2 sin θ cos θ, but it is intended for the sine rule to be used. Using the sine rule, I have got as far as

p / sin 2θ = q / sin θ

but I can't get any further. Help much appreciated!
How do p and q relate to P, Q, and R?

Thank you!

Eliz.
 

blamocur

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Oct 30, 2021
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Assuming that [imath]p[/imath] and [imath]q[/imath] are the lengths of the sides opposite to P and Q respectively, here is a hint:

Define point S on PQ so that PR = SR (i.e., triangle PRS is isosceles) , then consider the SQR triangle.
 

Dr.Peterson

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Nov 12, 2017
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Hi - could someone help me with this, please:

In a triangle PQR, angle PQR = θ and angle QPR = 2θ. Prove that cos θ = p/(2q).

This is meant to be solved without using the trig identity sin 2θ = 2 sin θ cos θ, but it is intended for the sine rule to be used. Using the sine rule, I have got as far as

p / sin 2θ = q / sin θ

but I can't get any further. Help much appreciated!
You're almost there!

What is the identity for sin(2θ)?
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
14,681
Hi - could someone help me with this, please:

In a triangle PQR, angle PQR = θ and angle QPR = 2θ. Prove that cos θ = p/(2q).

This is meant to be solved without using the trig identity sin 2θ = 2 sin θ cos θ, but it is intended for the sine rule to be used. Using the sine rule, I have got as far as

p / sin 2θ = q / sin θ

but I can't get any further. Help much appreciated!
Clearly I failed to read the whole question.

But how do you know what is "meant" and "intended"? Can you quote the entire problem as given to you?

In effect, this problem derives the angle sum identity! Is the idea that that identity has not been taught yet, though the law of sines has, and someone is using this to prove the former? Anything you can say about the context may help.

But I have found a solution, which turns out to be equivalent to what @blamocur must have in mind. I did it by constructing the perpendicular bisector of QR ...
 
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