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About Sherman D.
Expertise
I can answer questions dealing in mathematics of all kinds except for Physics and Calculus, but i can answer questions in Pre-Calculus and Chemistry. I can also answer questions in Recipes of all kinds. I can find games cheats/walkthroughs, but i can`t find a specific game online or offline. I can also do history and recipes for alcoholic beverages.

Experience
Mathematics, Recipes, History, and Games.

Education/Credentials
High School graduated. I graduated with honors, and i was in Beta Club for a year and a half.

Awards and Honors
Principle's list and A and B honor roll in high school only.

You are here: Experts > Science > Mathematics > Advanced Math > trigonometry

Topic: Advanced Math

Expert: Sherman D.
Date: 5/11/2008
Subject: trigonometry

Question
Solve the following right triangles: a. a = 117 ft, b = 16.35ft and b. B = 8* 29', a = 32.8 ft ?

Answer
117^2 + 16.35^2 = c^2
c = sqrt(13956.3225) or about 118.14ft

using this

C = 90, because its the right angle.

117^2 + (sqrt(13956.3225))^2 - 2(117sqrt(13956.3225))cos(B) = 16.35^2
B = about 7.955

16.35^2 + (sqrt(13956.3225))^2 - 2(16.35sqrt(13956.3225))cos(A) = 117^2

A = about 82.045

c = 118.14ft
A = 82.045°
B = 7.955°

--------------------------------------------

A = 180 - (90 + 8°29')
A = 180 - 98°29'
A = 81°31'

32.8/sin(81°31) = b/(sin(8°29))
b = (32.8sin(8°29))/sin(81°31)
b = 4.89ft

c = 32.8/sin(81°31)
c = 33.2

Just To Make Sure
4.89^2 + 33.2^2 - 2(4.89 * 33.2)cos(81°31') = 32.8^2

this works out, so

C = 90°
A = 81°31'
b = 4.89ft
c = 33.2ft

to find the exact value, you will have to work the them out.

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