Question

13 POINTS!!!

In ΔABC, c = 4.2 inches, ∠C=24° and ∠A=115°. Find the area of ΔABC, to the nearest 10th of an square inch.

Answer 1

12.9
`in^(2)`

Explanation:

So to find the area of this triangle, you will need to use the equation

Area =
`(1)/(2)`c*b*sin(A) =
`(1)/(2)`a*b*sin(C)

Here, we have ∠A, ∠C, and side c

We can use the fact that
`(a)/(sin(A)) = (b)/(sin(B)) = (c)/(sin(C))` so solve for the other variables we do not have.

First we can find the other angle B. Since ∠A + ∠B + ∠C = 180°,

∠B = 180° - ∠A - ∠C, which is ∠B = 180° - 115° - 24° = 41°

Now that we have all three angles, we can solve for the sides

Since we only have side c, we will manipulate the equation with c and one of the others to solve for a or b. Let's solve for side b first.

Since
`(b)/(sin(B)) =(c)/(sin(C))`, solving for b would give us
`b=(csin(B))/(sin(C))`. Then plugging in our values we get
`(4.2sin(41))/(sin(24))`= 6.77 = b

Now we can solve for the remaining side, a, using the same method.

Since
`(a)/(sin(A)) =(b)/(sin(B))`, solving for a would give us
`a=(bsin(A))/(sin(B))`. Then plugging on our values we get
`(6.77sin(115))/(sin(41))`= 9.36 = a

Now that we have all our angles and sides, we can plug in our numbers to either of our area equations ⇒

Area =
`(1)/(2)`c*b*sin(A)=
`(1)/(2)`(4.2)(6.77)sin(115) = 12.9
`in^(2)` or
`(1)/(2)`a*b*sin(C) =
`(1)/(2)`(9.36)(6.77)sin(24) = 12.9
`in^(2)`