# Trigonometry help?

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• Dec 8th 2008, 07:29 PM
4AM
Quote:

Originally Posted by 11rdc11
From there either use law of sines or sin or cos to TU

The laws? I haven't been taught the laws of sines, sin, or cos yet. Is there another way to solve the problem?
• Dec 8th 2008, 07:30 PM
tjsfury
Quote:

Originally Posted by 4AM
Wait, I don't get why we have to use cos?

Because the 50 is your Adjacent Side not Hypot as I said (look at the angles to see which side is the hypot)

Im going to go look at my thread now maybe somones helped me on it take a look if you know statistics and probability.
http://www.mathhelpforum.com/math-he...tribution.html
• Dec 8th 2008, 07:36 PM
4AM
Quote:

Originally Posted by tjsfury
Because the 50 is your Adjacent Side not Hypot as I said (look at the angles to see which side is the hypot)

Im going to go look at my thread now maybe somones helped me on it take a look if you know statistics and probability.
http://www.mathhelpforum.com/math-he...tribution.html

So the two triangles must be split up separately? I thought the 50 would the opposite side?

And sorry, I don't know enough about the statistics and probability unit enough to help you with your question. I can hardly understand my current math teacher right now.
• Dec 8th 2008, 07:39 PM
11rdc11
Quote:

Originally Posted by 4AM
So the two triangles must be split up separately? I thought the 50 would the opposite side?

And sorry, I don't know enough about the statistics and probability unit enough to help you with your question. I can hardly understand my current math teacher right now.

Yep the triangles have to be split
• Dec 8th 2008, 07:42 PM
Soroban
Hello, 4AM!

Quote:

$PQ = 50,\;\;\angle PSR = 32^o,\;\;\angle QSR = 216^o.\;\;\text{Find }RS.$
Code:

  -    P *   :      |  *   :      |      *   :    x |          *   :      |              *  50      |              32°  *   :    R * - - - - - - - - - - - * S   :      |            26°  *   :  50-x |          *   :      |    *   -    Q *

Let $x \:=\:PR \quad\Rightarrow\quad 50-x\:=\:RQ$

In right triangle $PRS\!:\;\;\tan32^o \:=\:\frac{x}{RS} \quad\Rightarrow\quad x \:=\:RS\!\cdot\!\tan32^o\;\;{\color{blue}[1]}$

In right triangle $QRS\!:\;\;\tan26^o \:=\:\frac{50-x}{RS} \quad\Rightarrow\quad x \:=\:50-RS\!\cdot\!\tan26^o\;\;{\color{blue}[2]}$

Equate [1] and [2]: . $RS\!\cdot\tan32^o \:=\:50 - RS\!\cdot\!\tan26^o \quad\Rightarrow\quad RS\!\cdot\!\tan32^o + RS\!\cdot\!\tan26^o \:=\:50$

Factor: . $RS\!\cdot\!(\tan32^o+\tan26^o) \:=\:50 \quad\Rightarrow\quad RS \:=\:\frac{50}{\tan32^o+\tan26^o}$

Therefore: . $RS \;\approx\;44.94$

Quote:

$\angle STW = 38.8^o,\;\;\angle STV = 53.8^o,\;\;VW = 12.5$ . Find $TU.$
Code:

    T * - - - - - - - S       | * *  38.8°       |  *15°*       |    *    *       |      *      *       |        *        *       |          *          *       |      53.8° *        38.8° *       * - - - - - - - * - - - - - - - *       U              V    12.5    W

We have: . $\angle STW = 38.8^o \quad\Rightarrow\quad \angle TWU = 38.8^o$
Since $\angle STV = 58.8^o$, then $\angle WTV = 15^o\:\text{ and }\:\angle TVU = 53.8^o$

In $\Delta TVW$, use the Law of Sines:

. . $\frac{TV}{\sin38.8^o} \:=\:\frac{12.5}{\sin15^o} \quad\Rightarrow\quad TV \:\approx\:30.36$

In right triangle $TUV\!:\;\;\sin53.8^o \:=\:\frac{TU}{TV} \quad\Rightarrow\quad TU \:=\:30.26\!\cdot\!\sin53.8^o$

Therefore: . $TU \;\approx\;24.42$

• Dec 8th 2008, 07:42 PM
4AM
Quote:

Originally Posted by 11rdc11
Yep the triangles have to be split

If they're split though, why is the side still 50 m? Wouldn't it be smaller?
• Dec 8th 2008, 07:44 PM
11rdc11
Side PQ and QS are both 50
• Dec 8th 2008, 07:49 PM
4AM
Quote:

Originally Posted by 11rdc11
Side PQ and QS are both 50

How do you know that though? Sorry, these questions probably seem really stupid to you but I don't really understand how to solve these questions.

Thank you for your help though! =)
• Dec 8th 2008, 08:06 PM
11rdc11
Quote:

Originally Posted by 4AM
How do you know that though? Sorry, these questions probably seem really stupid to you but I don't really understand how to solve these questions.

Thank you for your help though! =)

Because the triangle is an isoceles so two sides are the same
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