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

Answer:

12.9 $in^{2}$

Step-by-step explanation:

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

Area = $\frac{1}{2}$c*b*sin(A) = $\frac{1}{2}$a*b*sin(C)

Here, we have ∠A, ∠C, and side c

We can use the fact that $\frac{a}{sin(A)} = \frac{b}{sin(B)} = \frac{c}{sin(C)}$ 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 $\frac{b}{sin(B)} =\frac{c}{sin(C)}$, solving for b would give us $b=\frac{csin(B)}{sin(C)}$. Then plugging in our values we get $\frac{4.2sin(41)}{sin(24)}$= 6.77 = b

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

Since $\frac{a}{sin(A)} =\frac{b}{sin(B)}$, solving for a would give us $a=\frac{bsin(A)}{sin(B)}$. Then plugging on our values we get $\frac{6.77sin(115)}{sin(41)}$= 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 =$\frac{1}{2}$c*b*sin(A)= $\frac{1}{2}$(4.2)(6.77)sin(115) = 12.9$in^{2}$ or $\frac{1}{2}$a*b*sin(C) = $\frac{1}{2}$(9.36)(6.77)sin(24) = 12.9$in^{2}$