Question

Given that ABC has m C = 88°, a = 12, and b = 22, find the remaining side length c and angles A and B, rounded to the nearest tenth.

Answer 1

We can solve this problem by using the law of cosines:

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

The given information is:

C=88

a=12

b=22

Then:

$\begin{gathered} c^2=12^2+22^2-2(12)(22)\cos 88 \\ c^2=144+484-528\cdot0.035 \\ c^2=144+484-18.43 \\ c^2=609.57 \\ c=\sqrt[]{609.57} \\ c=24.7 \end{gathered}$

Now, we can use the law of sines to find the other angles:

$\frac{\sin A}{a}=\frac{\sin C}{c}$

Replace the values and solve for A:

$\begin{gathered} \sin A=\frac{a\sin C}{c} \\ \sin A=\frac{12\cdot\sin 88}{24.7} \\ \sin A=\frac{12}{24.7} \\ \sin A=\text{0}.486 \\ A=\sin ^{-1}(0.486) \\ A=29.1 \end{gathered}$

And the sum of the interior angles of a triangle is 180°, then:

$\begin{gathered} A+B+C=180\degree \\ 29.1\degree+B+88\degree=180\degree \\ B=180\degree-29.1\degree-88\degree \\ B=62.9\degree \end{gathered}$