• Jan 16th 2006, 09:46 AM
tristan1

http://img388.imageshack.us/img388/7417/helphelp0ag.jpg
• Jan 16th 2006, 10:11 AM
earboth
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
• Jan 16th 2006, 10:40 AM
tristan1
Thanks and they are rectangular
• Jan 16th 2006, 12:13 PM
MathGuru
The correct english for this type of triangle is "right triangle" which implies 1 of the angles is 90 degrees.

Thanks Earboth
• Jan 17th 2006, 09:02 AM
tristan1

Top left: 38.60m
Top right: 5.65m

Bottom left: 171.334m
Bottom right: 45.583m
• Jan 24th 2006, 10:54 AM
earboth
Quote:

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