You need not draw any circles! The problem is quite simple.
According to sine rule:
We have : .
Using double angle formula we get :..
Doing some self-teach on Maths, and got stumped by the following:-
In triangle PQR, angle PQR = theta and angle QPR = 2 theta. Prove cos (theta) = p / 2q
This question occurs before presentation of trignometric identities, so it implies that knowledge of these is not required to solve the above.
The question is given as part of a discussion of the Sine Rule, given in the form
a / sin A = b / sin B = c / sin C
I've perhaps gone down the wrong the wrong route by:-
- Making P the centre of a circle
- R lie on the circumference of a circle
- Q a point in the circle (and perhaps on)
- Drawing a line of length p from R to interesct the circle, drawing a diameter of length 2q from R to P and onto another point, giving a right angle triangle with a hypotenuse 2q and base p, with an angle whose cosine is p / 2q
What I cannot work out is a way to relate the angle of the triangle I constructed to the angles given in the original triangle.
( The text derives the Sine rule from i) angle in a semicircle, ii) angles in same segment, iii) opposite angles of a cyclic quadrilateral).
Knowing where I'm going wrong with this would be most helpful. Thanks!
Thanks for the reply.
The double angle formula does get the desired result.
The difficulty with self teach is of course that it is not always clear what the 'right' way to solve the problem is. The text I am working with had not introduced the sin (2 theta ) = 2 sin(theta) cos (theta). So I'm unsure whether that was a 'legitimate' way of proving the result.