Question

In △ABC, m∠A=39°, a=11, and b=13. Find c to the nearest tenth.

Answer 1

For this problem, we are going to use the law of sines, which states:

$\dfrac{\sin{A}}{a} = \dfrac{\sin{B}}{b} = \dfrac{\sin{C}}{c}$

In this case, we have an angle and two sides, and we are trying to look for the third side. First, we have to find the angle which corresponds with the second side, $B$. Then, we can find the third side. Using the law of sines, we can find:

$\dfrac{\sin{39^{\circ}}}{11} = \dfrac{\sin{B}}{13}$

We can use this to solve for $B$:

$13 \cdot \dfrac{\sin{39^{\circ}}}{11} = \sin{B}$

$B = \sin^{-1}{\Big(13 \cdot \dfrac{\sin{39^{\circ}}}{11}\Big)} \approx 48.1$

Now, we can find $C$:

$C = 180^{\circ} - 48.1^{\circ} - 39^{\circ} = 92.9^{\circ}$

Using this, we can find $c$:

$\dfrac{\sin{39^{\circ}}}{11} = \dfrac{\sin{92.9^{\circ}}}{c}$

$c = \dfrac{11\sin{92.9^{\circ}}}{\sin{39^{\circ}}} \approx \boxed{17.5}$

c is approximately 17.5.