Why is my sine greater than 1?

[bushra1175]

Hi everyone. I'm trying to work out the question below:


I started off by trying to work out angle C. Here is my working out using the law of sines:


I've looked over everything a dozen times and I'm pretty sure I'm using the law of sines correctly, but I know that sine C is not supposed to be larger than 1. This has given me an error on the calculator when trying to get the inverse of sine C. What am I doing wrong?

 

[Dr.Peterson]

I don't get 1.06. Did you forget to set your calculator to degrees?

Of course, I hope you know that it is possible for a triangle in such a problem not to exist; if 1.06 were the correct sine, then that is what it would be telling you.

 

[bushra1175]

I don't get 1.06. Did you forget to set your calculator to degrees?

Of course, I hope you know that it is possible for a triangle in such a problem not to exist; if 1.06 were the correct sine, then that is what it would be telling you.

Yes! Thank you so much!

 

[forestview]

I don't get 1.06 either, check your calcs again it's less than 1 (0.91826)

 

[Dr.Peterson]

I don't get 1.06 either, check your calcs again it's less than 1 (0.91826)

It looks like you missed my explanation of the error: bushra1175 had his calculator set to radian mode, which is why they got

[MATH]\frac{20\sin40 rad}{14} =1.064447...[/MATH] .​

Putting it in degree mode, you get

[MATH]\frac{20\sin40^\circ}{14} = 0.918268...[/MATH] as you say.​

This fully answered the question.