I am trying to find the area of a triangle..Given angle sides..?

Question

Find the area of the triangle given that...

c = 40 m
b = 20 m
A = 48 degrees

I need the answer in m^2
so its one of the following answers I dont know which one, or how to figure this type of problem out?
Multiple choice
a) 1000 m^2
b) 120 m^2
c) 80 m^2
d) 297 m^2

Answer 1

(1/2)40*20*sin(48)=297 m^2 (the square is just the unit for area)

Answer 2

There's a neat formula for this.

A = 1/2 a b sin c

In this case,

A = 1/2 (40) (20) sin 48° = 297 m²

The derivation of the formula is pretty simple if you draw the triangle.

Put angle A on the right, with the 40m side on the bottom (base of the triangle) and the 50m side sloping up to the left. Go ahead and draw the 3rd side connecting them.

You've already got the base (40m). Draw a line straight down from the point of the triangle to the base. That's the height, right?

The height is the "opposite" of the 48° angle, and the 20m side is the hypotenuse:

sin 48° = h/20
h = 20 sin 48°

And the area of a triangle is 1/2 b h, so that's:

1/2 (40) (20 sin 48°)

Answer 3

d) 297.26 square meters

Assume the 40 m side is the base. The height is the length of the other side, multiplied by the sine of 48 degrees. The area is one-half the product of the base and the height. So:

A = .5 x 40 x 20 x sin(48) = 297.257930...

Answer 4

area of triangle is = 0.5 cbsin A
= 0.5*40*20*sin 48 =297 m^2

Answer 5

Using the laws of cosine, get a:

a^2 = b^2 + c^2 - 2bc cos(A)
a = 30.49

Using Heron's Formula:
let s = (a + b + c)/2
A = sqrt (s * (s-a) * (s-b) * (s-c))
A = 297.31 m^2