trigo problem?

Question

Solve for the unknown parts and find its area
a= 23 b=43 c=31
(Find the three angles and the area)

Answer 1

a ² = b ² + c ² - 2 b c cos A
cos A = (b ² + c ² - a ²) / (2 b c)
cos A = (43² + 31² - 23²) / (2 x 43 x 31)
A = 31.2°

23/sin 31.2° = 43 / sin B
sin B = 43 sin 31.2° / 23
B = 75.4°

C = 180° - 75.4° - 31.2°
C = 73.4°

Area = (1/2) a b sin C
Area = (1/2) (23) (43) sin 73.4°
Area = 473.9 units ²

Answer 2

The angles can be found by use of the law of cosines.

This states that
cos(C) = (a^2 + b^2 - c^2)/2ab

so

cos(C) = (23^2 + 43^2 - 31^2)/(2*23*43)
= (529 + 1849 - 961) / 1978
= 1417 / 1978
= 0.716

C = 44.244

Likewise
cos(A) = (43^2 + 31^2 - 23^2)/(2*43*31)
= 2281 / 2666
= 0.855
A = 31.175

and

cos(B) = (23^2 + 31^2 - 43^2)/(2*23*31)
= -359 / 1426
= -.252
B = 104.581

The area of a triangle can be found by the formula
A = sqrt(s(s-a)(s-b)(s-c)) where s = (a+b+c)/2

s = (23 + 43 + 31)/2
= 48.5

so A = sqrt(48.5*(48.5-23)(48.5-43)(48.5-31))
= sqrt(48.5*25.5*5.5*17.5)
= sqrt(119037.1875)
= 345.018

Answer 3

I take it these are the 3 lengths of a triangle.

First use cosine rule

a^2 = b^2 + c^2 -2bcCosA

where angle A is opposite side a

to find the other 2 angles use this rule again or the sine rule:

a/sinA = b/sinB = c/sinC

then area of circle = 1/2 absinC

Answer 4

Use a law of Cosines to find the first angle. Then law of sines to find the second, finally subtract the two from 180 to get the third.
Area formulas are abundant once you get these. But you can use the three sides given in
sqrt[s(s-a)(s-b(s-c)] where s = (a+b+c)/2