# Math Help - Help Please!

2. Hello tristan1,
I presume that your triangles are rectangular.

So use the equations which define the sin- and cos-function:

$\sin{52^\circ}=\frac{32}{x}$

$\cos{23^\circ}=\frac{5.2}{x}$

$\sin{71^\circ}=\frac{162}{x}$

$\sin{71^\circ}=\frac{431}{x}$

I guessed the last number only, so may be you've to repair my equation a little bit.

To solve the equations should be of no trouble for you.

Bye

3. Thanks and they are rectangular

4. The correct english for this type of triangle is "right triangle" which implies 1 of the angles is 90 degrees.

Thanks Earboth

Top left: 38.60m
Top right: 5.65m

Bottom left: 171.334m
Bottom right: 45.583m

6. Originally Posted by tristan1

Top left: 38.60m
Top right: 5.65m

Bottom left: 171.334m
Bottom right: 45.583m
Hello,

you did a very fine job! The only thing I've to apologize is I've made a mistake in my first reply.

The equation for the bottom right question is:
$\cos 71^\circ = \frac{43.1}{x} \ \Rightarrow \ x=\frac{43.1}{\cos 71^\circ} \approx 132.38$

Best wishes to you

Earboth