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This question has me absolutely stumped. This is the image of the question, how can I work out $x$? I've been doing a variety of attempts but I just cant get it.

This question has me absolutely stumped. This is the image of the question, how can I work out $x$? I've been doing a variety of attempts but I just cant get it.

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You can use the Law of Sines to find one of the other side lengths. Then it should be an easy matter to find $x$ using one of the trig functions. –  David Mitra Feb 13 at 9:46

2 Answers 2

Another approach is to note that $$ x\cot(33^\circ)+x\cot(25^\circ)=20 $$ to get $$ \begin{align} x &=\frac{20}{\cot(33^\circ)+\cot(25^\circ)}\\[4pt] &\approx5.42833368289824 \end{align} $$

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Call the side opposite $33^\circ$ as $a$. Therefore we have: $$\dfrac {x}{a}=\sin 25^\circ$$

and from the sine rule for the triangle we know that: $$\dfrac {20}{\sin 122^\circ}=\dfrac {a}{\sin 33^\circ}$$

Therefore from the above two equations we have $x=\dfrac{20\times\sin 33^\circ \times \sin 25^\circ}{\sin 122^\circ} $, or $$x\approx5.428336828982414$$

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To compare our answers: $$\begin{align}\frac1{\cot(33^\circ)+\cot(25^\circ)} &=\frac{\sin(33^\circ)\sin(25^\circ)}{\cos(33^\circ)\sin(25^\circ)+\cos(25^\circ‌​)\sin(33^\circ)}\\[4pt] &=\frac{\sin(33^\circ)\sin(25^\circ)}{\sin(58^\circ)}\\[4pt] &=\frac{\sin(33^\circ)\sin(25^\circ)}{\sin(122^\circ)}\end{align}$$ –  robjohn Feb 13 at 10:17
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@robjohn, both of us have tried different things which are essentially the same... The question was quite elementary though... –  Apurv Feb 13 at 10:39
    
Agreed. I was just showing how the answers were the same even though they look quite different. –  robjohn Feb 13 at 11:14

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