SOLUTION: I have a question, I was wondering if you could help me with how to find the angles of a triangle that has lengths of 11cm 8cm and 5cm. I am unsure of how to find the angles of the

Question 1030595: I have a question, I was wondering if you could help me with how to find the angles of a triangle that has lengths of 11cm 8cm and 5cm. I am unsure of how to find the angles of the triangles with only having the lengths given how could I do that? I think that I can find it without using a protractor as it is online but I am unsure could you help me find the angles of the triangle please
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
look at the diagram.
see the comments below.

$$$

set your triangle up as follows:

c = side opposite angle C
b = side opposite angle B
a = side opposite angle A

the law of cosines allows you to find the angles given the sides.

that law states:

c^2 = a^2 + b^2 - 2ab*cos(C)

a^2 = b^2 + c^2 - 2bc*cos(A)

b^2 = a^2 + c^2 - 2ac*cos(B)

you would need to solve for two of the angles and then you can get the third angle automatically by subtracting the sum of those two angles from 180.

i'll solve for angle C and for angle A.

for angle C, start with:

c^2 = a^2 + b^2 - 2ab*cos(C)

solve for cos(C) to get:

cos(C) = (a^2 + b^2 - c^2) / (2*a*b)

replace letters with values to get:

cos(C) = (8^2 + 11^2 - 5^2) / (2*8*5)

simplify to get:

cos(C) = .9090909091

solve for C to get:

C = arccos(.9090909091) = 24.61997733 degrees.

for angle A, start with:

a^2 = b^2 + c^2 - 2bc*cos(A)

solve for cos(A) to get:

cos(A) = (b^2 + c^2 - a^2) / (2*b*c)

replace letters with values to get:

cos(A) = (11^2 + 5^2 - 8^2) / (2*11*5)

solve for cos(A) to get:

cos(A) = .7454545455

solve for A to get:

A = arccos(.7454545455) = 41.80184419 degrees.

now that you have angle A and angle C, you can solve for angle B by subtracting their sum from 180.

angle B = 180 - 24.61997733 - 41.80184419 = 113.5781785 degrees.

you could also solve for angle B using the law of cosines again.

you would get the same answer.

here's a lesson on law of sines and law of cosines that can help you understand what the formulas are and how to use them.

http://www.regentsprep.org/regents/math/algtrig/ATT12/indexATT12.htm

select law of sines lesson and then select law of cosines lesson.

here's another reference talking about the same thing.

http://www.mathwarehouse.com/trigonometry/law-of-sines-and-cosines.php

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