The laws? I haven't been taught the laws of sines, sin, or cos yet. Is there another way to solve the problem?

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- Dec 8th 2008, 07:29 PM4AM
- Dec 8th 2008, 07:30 PMtjsfury
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 PM4AM
- Dec 8th 2008, 07:39 PM11rdc11
- Dec 8th 2008, 07:42 PMSoroban
Hello, 4AM!

Quote:

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

: | *

: | *

: x | *

: | *

50 | 32° *

: R * - - - - - - - - - - - * S

: | 26° *

: 50-x | *

: | *

- Q *

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

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

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

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

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

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

Quote:

$\displaystyle \angle STW = 38.8^o,\;\;\angle STV = 53.8^o,\;\;VW = 12.5$ . Find $\displaystyle TU.$Code:`T * - - - - - - - S`

| * * 38.8°

| *15°*

| * *

| * *

| * *

| * *

| 53.8° * 38.8° *

* - - - - - - - * - - - - - - - *

U V 12.5 W

We have: .$\displaystyle \angle STW = 38.8^o \quad\Rightarrow\quad \angle TWU = 38.8^o$

Since $\displaystyle \angle STV = 58.8^o$, then $\displaystyle \angle WTV = 15^o\:\text{ and }\:\angle TVU = 53.8^o$

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

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

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

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

- Dec 8th 2008, 07:42 PM4AM
- Dec 8th 2008, 07:44 PM11rdc11
Side PQ and QS are both 50

- Dec 8th 2008, 07:49 PM4AM
- Dec 8th 2008, 08:06 PM11rdc11