# Trigonometry help?

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• Dec 8th 2008, 06:34 PM
4AM
Trigonometry help?
How would you find the length of RS? What do you have to find first? Can someone please explain this to me step by step? Help is appreciated!

http://img355.imageshack.us/img355/7590/pg3367ut0.jpg

Same for this one, how would you find the length of TU? Where would you start and what do you have to do?
http://img247.imageshack.us/img247/7826/pg3368qt4.jpg
• Dec 8th 2008, 06:49 PM
tjsfury
Problem #1
$\displaystyle \sin$ = $\displaystyle \frac{opposite}{hyp}$
If you use the angle with 32 deg you would have Opposite = 25 and
Hyp = x

when you get that you can use a$\displaystyle ^2$ + b$\displaystyle ^2$ = c$\displaystyle ^2$

A = 25
B = RS
C = the hyp you found earlier

Problem #2

Im not sure on the second one because I don't know if you can use S$\displaystyle \frac{o}{h}$ C$\displaystyle \frac{A}{H}$ T$\displaystyle \frac{O}{A}$ or not without a Right angle its been too long for me if i have time ill figure out how to do it and help you out.
• Dec 8th 2008, 06:58 PM
4AM
Quote:

Originally Posted by tjsfury
Problem #1
$\displaystyle \sin$ = $\displaystyle \frac{opposite}{hyp}$
If you use the angle with 32 deg you would have Opposite = 25 and
Hyp = x

when you get that you can use a$\displaystyle ^2$ + b$\displaystyle ^2$ = c$\displaystyle ^2$

A = 25
B = RS
C = the hyp you found earlier
Problem #2 ---- give me a second to look at it

I don't think the opposite side of the triangle (if you're looking through at 32 degrees) would be 25 though. The base of the triangles aren't the same length, but together they add up to 50 meters.

Edited:
One of triangle's bases has a bigger length than the other (for the first question)
• Dec 8th 2008, 07:00 PM
tjsfury
Yea i agree let me get a pencil out and think about it im bored enough to help lol
• Dec 8th 2008, 07:02 PM
4AM
Quote:

Originally Posted by tjsfury
Yea i agree let me get a pencil out and think about it im bored enough to help lol

Haha, alright. I appreciate you helping me =)
• Dec 8th 2008, 07:06 PM
tjsfury
ok i figured it out, i think..... 32+26 = 58 (1 angle of big triangle)
180 - (32 + 90) = 58 (angle of top side) 2 angles are the same 2 opposite sides are the same in length
.... been a long time all angles of a triangle added together = 180 right?
• Dec 8th 2008, 07:08 PM
11rdc11
Quote:

Originally Posted by 4AM
How would you find the length of RS? What do you have to find first? Can someone please explain this to me step by step? Help is appreciated!

http://img355.imageshack.us/img355/7590/pg3367ut0.jpg

Same for this one, how would you find the length of TU? Where would you start and what do you have to do?
http://img247.imageshack.us/img247/7826/pg3368qt4.jpg

The 1st triangle is an iscolses triangle. The top left angle is 58 degrees and the bottom left is 64 degrees, and the middle right is 58 degrees

So $\displaystyle \cos {26} = \frac{RS}{50}$
• Dec 8th 2008, 07:09 PM
tjsfury
ok yea i looked it up all angles = 180

So your bottom leg SQ = 50m

Because of the equal angles then you can use..... Sin with 26 knowing that H = 50
• Dec 8th 2008, 07:10 PM
tjsfury
Quote:

Originally Posted by 11rdc11
The 1st triangle is an iscolses triangle. The top left angle is 58 degrees and the bottom left is 64 degrees, and the middle right is 58 degrees

Lol i figured that out right before you posted it :] so how about his second one?

for some reason i remember something about having a line like a Z shape inner and outter angles being equal....
in his problem would the bottom right angle be 38.8? Probably not
• Dec 8th 2008, 07:13 PM
11rdc11
Quote:

Originally Posted by tjsfury
ok yea i looked it up all angles = 180

So your bottom leg SQ = 50m

Because of the equal angles then you can use..... Sin with 26 knowing that H = 50

U mean cos not sin right(Wink)
• Dec 8th 2008, 07:17 PM
tjsfury
Quote:

Originally Posted by 11rdc11
U mean cos not sin right(Wink)

Yea i do I was looking at the 90 degree mark and didn't look at the angles (if you look i said that H=50.... yea that side isn't H its A) what about his #2? any ideas
• Dec 8th 2008, 07:21 PM
11rdc11
Quote:

Originally Posted by tjsfury
Yea i do I was looking at the 90 degree mark and didn't look at the angles (if you look i said that H=50.... yea that side isn't H its A) what about his #2? any ideas

Yea you have to find all the angles and then use the law of sines

$\displaystyle \frac{\sin{15}}{12.5} = \frac{\sin{38.8}}{TV}$
• Dec 8th 2008, 07:25 PM
tjsfury
Ha I was right about the 38.8 angle.... Z angles (alternate angles) are equal.... shouldn't be too hard knowing that
Angles
• Dec 8th 2008, 07:27 PM
11rdc11
From there either use law of sines or sin or cos to TU
• Dec 8th 2008, 07:28 PM
4AM
Quote:

Originally Posted by 11rdc11
The 1st triangle is an iscolses triangle. The top left angle is 58 degrees and the bottom left is 64 degrees, and the middle right is 58 degrees

So $\displaystyle \cos {26} = \frac{RS}{50}$

Wait, I don't get why we have to use cos?
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