With question 1 I have guessed the unknown side lengths using common pythagorean triples. Is there another way to get them? I have done 2a. Can I have a clue for 2b? I expect it uses addition formulae.
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Originally Posted by Stuck Man With question 1 I have guessed the unknown side lengths using common pythagorean triples. Is there another way to get them? I have done 2a. Can I have a clue for 2b? I expect it uses addition formulae. 2(b) can be calculated using $\displaystyle Sin(A-B)=SinACosB-CosASinB$ given that $\displaystyle A=180^o$ and $\displaystyle B=2x$ and $\displaystyle Sin180^0=0$ and $\displaystyle Cos180^o=-1$
Thanks. I copied the diagram for question 1 and missed the length of QP so theres no need to reply to that now.
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