# B=20° C=75° b=5 solve the triangle?

## When I set up the equation using law of sines it looked like this 5/sin(20)=x/sin(75) then x=-2.124 what did I do wrong?

Feb 28, 2018

$\angle A = {85}^{\circ}$
$a \approx 14.56$
$c \approx 14.12$

#### Explanation:

You're calculator is in radians! That's why. It should be in degrees.

You are on the right track however!

Let's draw the triangle [Note: This is not to scale]:

The missing angle can be found by knowing that the sum of the internal angles of any triangle is always equal to ${180}^{\circ}$

So $\angle A = {85}^{\circ}$

The law of sines states:

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

We know:

$b = 5$
$\angle A = {85}^{\circ}$
$\angle B = {20}^{\circ}$
$\angle C = {75}^{\circ}$

What we want to know is

a=?
c=?

We can find the missing sides using the law of sines. So to find $a$

$\frac{a}{\sin} A = \frac{b}{\sin} B$

Pluggin in what we know:

$\frac{a}{\sin} \left({85}^{\circ}\right) = \frac{5}{\sin} \left({20}^{\circ}\right)$

$\frac{a}{\sin} \left({85}^{\circ}\right) = 14.619022$

$a = \sin \left({85}^{\circ}\right) \cdot 14.619022$

color(red)(a~~14.56

Similarly for $c$:

$\frac{b}{\sin} B = \frac{c}{\sin} C$

$\frac{5}{\sin} \left({20}^{\circ}\right) = \frac{c}{\sin} \left({75}^{\circ}\right)$

$14.619022 = \frac{c}{\sin} \left({75}^{\circ}\right)$

$\sin \left({75}^{\circ}\right) \cdot 14.619022 = c$

color(red)(14.12~~c