Really stuck with these.. please help!

Results 1 to 6 of 6

- Jan 16th 2006, 09:46 AM #1

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- Nov 2005
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- Jan 16th 2006, 10:11 AM #2
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 AM #3

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- Nov 2005
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- 12

- Jan 16th 2006, 12:13 PM #4

- Jan 17th 2006, 09:02 AM #5

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- Nov 2005
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- Jan 24th 2006, 10:54 AM #6Originally 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