Really stuck with these.. please help!

http://img388.imageshack.us/img388/7417/helphelp0ag.jpg

Printable View

- Jan 16th 2006, 09:46 AMtristan1Help Please!
Really stuck with these.. please help!

http://img388.imageshack.us/img388/7417/helphelp0ag.jpg - Jan 16th 2006, 10:11 AMearboth
Hello tristan1,

I presume that your triangles are rectangular.

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

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

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

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

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

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 AMtristan1
Thanks and they are rectangular

- Jan 16th 2006, 12:13 PMMathGuru
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 AMtristan1
I've had a go, please check my answers someone!

Top left: 38.60m

Top right: 5.65m

Bottom left: 171.334m

Bottom right: 45.583m - Jan 24th 2006, 10:54 AMearbothQuote:

Originally Posted by**tristan1**

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:

$\displaystyle \cos 71^\circ = \frac{43.1}{x} \ \Rightarrow \ x=\frac{43.1}{\cos 71^\circ} \approx 132.38 $

Best wishes to you

Earboth