# Math Help - Trigonometry help?

1. 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?

2. 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

3. 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.

4. 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

5. Hello, 4AM!

$PQ = 50,\;\;\angle PSR = 32^o,\;\;\angle QSR = 216^o.\;\;\text{Find }RS.$
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: . $RS \;\approx\;44.94$

$\angle STW = 38.8^o,\;\;\angle STV = 53.8^o,\;\;VW = 12.5$ . 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: . $TU \;\approx\;24.42$

6. 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?

7. Side PQ and QS are both 50

8. 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! =)

9. 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

Page 2 of 2 First 12