trig angles of a triangle, please help!?

Question

If a=5, b=8, and c=9

find alpha & write your answer in terms of an inverse trig function.


can someone please help me and explain this to me, I am completely lost and I'm studying for my final, if you could help I would appreciate it!! thanks

Answer 1

Presuming that α is the angle opposite side a...
Law of Cosines
a² = b² + c² - (2bc)cos(α)

Do the algebra and eventually you're going to get
something=cos(α)

I think that the "something" is: (b² + c² - a²)/(2bc) but that's in my head, so you'll want to do it yourself to be sure.

Next, plug in the values of a, b, and c into whatever you get, and do the arithmetic. You'll get some value.

The inverse function you're looking for is
arccos(some value)=α

You can find the other angles the same way.

Answer 2

Well, I'm going to assume that your triangle is not a right triangle, since a^2 + b^2 =/= c^2. In other words, 25 + 64 = 89, and the square root of 89 isn't 9, so it's not a right triangle. Instead, this is an oblique triangle. To find the angles of an oblique triangle, use the law of cosines: a^2 = b^2 + c^2 - 2bc*cosA.
In other words, side a squared = side b squared plus side c squared - 2 x side b x side c x cos angle A.

Answer 3

Use the Law of Cosines.

Find cosA.

a² = b² + c² - 2bc cosA
2bc cosA = b² + c² - a²

cosA = (b² + c² - a²) / (2bc) = (8² + 9² - 5²) / (2*8*9)
cosA = (64 + 81 - 25) / 144 = 120/144 = 5/6

A = arccos(5/6)

Find B the same way.

Then C = 180 - A - B.

Answer 4

You can use the law of sins or cos. Arc tan of 8/5 gives you an angle. c. c/sinc=b/sinb=a/sina.