i need help/ trigonometry plz?

Question

Note that the point value of each question is given in parentheses

1. (10) Solve the triangle for which the given parts are a= 27, b= 21 and c= 24

Answer 1

Law of Cosines

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

27^2 = 21^2 + 24^2 - 2(21 * 24)cos(A)
729 = 441 + 576 - 2(504)cos(A)
729 = 1017 - 1008cos(A)
-288 = -1008cosA
2/7 = cosA
A = 73°23'54.42"

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b^2 = a^2 + c^2 - 2ac(cos(B))
21^2 = 27^2 + 24^2 - 2(27 * 24)cos(B)
441 = 729 + 576 - 2(648)cosB
441 = 1305 - 1296cosB
-864 = -1296cosB
cosB = (2/3)
B = 48°11'22.87"

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c^2 = a^2 + b^2 - 2ab(cosC)
24^2 = 27^2 + 21^2 - 2(27 * 21)cosC
576 = 729 + 441 - 2(567)cosC
576 = 1170 - 1134cosC
-1134cosC = -594
cosC = 11/21
C = 58°24'42.71"

ANS :
A = 73°23'54.42"
B = 48°11'22.87"
C = 58°24'42.71"

Answer 2

I agree.

You already have the sides, so, you have to calculate the angles.

Just do this:

a^2 = b^2+c^2 - 2bc cos A

2bc cos A = b^2 + c^2 - a^2

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

You have a, b and c, calculate cos A, this is a number.

Then find cos^(-1) from this number, this will be A

You can calcualte the other angles using a similar formulas:

b^2 = a^2 + c^2 - 2ac cos B

and

c^2 = a^2 + b^2- 2ab cos C

Ana

Answer 3

Sounds like a job for the Law of Cosines, a^2 = b^2 + c^2 -2bc cos A.