1. ## Sine Rule Help please

Hi maths gurus.

I'm in need of some help here with the sine rule.

I have my sine rule values 3.1/sin20 = 5/sineB = 6/SineC

I am trying to find the value for the sine of B so what I did was 3.1/Sine(20)*5 to get 1.81 If I do the inverse sine it comes up with an error on my calculator so is this what I'm meant to do or what?

2. ## Re: Sine Rule Help please

You want $sin(B) = \frac{5sin(20)}{3.1}$

$\frac{3.1}{5sin(20)}$ will never work, as it is greater than 1.

3. ## Re: Sine Rule Help please

Ah Okay thank you very much! It worked

4. ## Re: Sine Rule Help please

A little more clarification: you started from $\frac{3.1}{sin(20)}= \frac{5}{sin(B)}$ and divided both sides by 5 to get
$\frac{3.1}{5 sin(20)}= \frac{1}{sin(B)}$. The right side is $\frac{1}{sin(B)}$, NOT sin(B) so you cannot immediately take the arcsine. To solve for B, you have to take the reciprocal of both sides to get $sin(B)= \frac{5 sin(20)}{3.1}$. then take the arcsine.