Question

I need help figuring out how to find sides a and b using the law of sine

Answer 1

Given the triangle ABC below.

a is the side facing b is the side facing

c is the side facing

We ara interested in calculating the value of side a and b.

To do this, we need to apply the "sine rule"

Sine rule state that

`(a)/(\sin A)=(b)/(\sin B)=(c)/(\sin C)`

Where

a is the side facing b is the side facing

c is the side facing

To calculate b,

B = 95 , b = ?

C = 48, c=100

`\begin{gathered} (b)/(\sin B)=(c)/(\sin C) \\ (b)/(\sin 95)=\frac{100}{\sin \text{ 48}} \\ \\ b\text{ x sin48=100 x sin95} \\ b=\frac{100\text{ x sin95}}{\sin 48} \\ b=134.05 \end{gathered}`

b = 134 ( to nearest whole number)

To calculate a:

A = 37, a = ?

C = 48, c=100

`\begin{gathered} (a)/(\sin A)=(c)/(\sin C) \\ (a)/(\sin37)=(100)/(\sin 48) \\ a\text{ x sin48 = 100 x sin37} \\ a=\frac{100\text{ x sin37}}{\sin 48} \\ a=80.98 \\ \end{gathered}`

a = 81 ( to the nearest whole number)