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!