Using the Law of Sines - Getting a Domain Error on my Calculator

[opus]

Homework Statement

Solve for leg C in the following picture

Relevant Equations

Law of Sines
Law of Cosines

So we'd like to find leg C.
But we can't use Law of Cosines yet so we will use Law of Sines.

We can easily find the length of A and this is $\sqrt{13}$.
With some geometry we can see that $\angle a = 53.1$.
We can now use Law of sines.

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

We want to find $\angle b$. So we have

$$\angle b = \sin^{-1}\left(\frac{B\sin(a)}{A}\right)$$

$$\angle b = \sin^{-1}\left(\frac{5\sin(53.1)}{\sqrt{13}}\right)$$

However this gives me a domain error in my calculator which makes sense because the argument is greater than 1. It's been awhile since I've done trig, so I'm wondering how to remedy this so that I can use the Law of Cosines after this?

Thanks

 

[mitochan]

Leg C and the dotted line in the figure do not seem parallel to me. 53.1 degree angle may be from x axis.

 

[SammyS]

Homework Statement:: Solve for leg C in the following picture
Relevant Equations:: Law of Sines
Law of Cosines

So we'd like to find leg C.
But we can't use Law of Cosines yet so we will use Law of Sines.

We can easily find the length of A and this is $\sqrt{13}$.
With some geometry we can see that $\angle a = 53.1$.
We can now use Law of sines.

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

We want to find $\angle b$. So we have

$$\angle b = \sin^{-1}\left(\frac{B\sin(a)}{A}\right)$$

$$\angle b = \sin^{-1}\left(\frac{5\sin(53.1)}{\sqrt{13}}\right)$$

However this gives me a domain error in my calculator which makes sense because the argument is greater than 1. It's been awhile since I've done trig, so I'm wondering how to remedy this so that I can use the Law of Cosines after this?

Thanks

It seems to me that simple geometry shows ∠a to be somewhat less than 53.1°.

This problem is much easier to do if you use vector components.

 

[ehild]

@SammyS is right. A,B,C are shown as vectors. A is given with its x,y components, B is given with its magnitude and with the angle it makes clockwise with the positive x axis. The x component of vector C is asked.
How is related vector C with vectors A and B?

 

[mitochan]

The figure shows
$$\mathbf{A}+\mathbf{B}=-\mathbf{C}$$

 

[ehild]

The figure shows
$$\mathbf{A}+\mathbf{B}=-\mathbf{C}$$

Correct. How do you add two vectors by components?
You know the components of A: (2,3). What are the components of B?

 

[mitochan]

$$(5 cos\theta, 5 sin\theta)$$ where $\theta$ =- 53.1 degree.

 

[ehild]

$$(5 cos\theta, 5 sin\theta)$$ where $\theta$ =
- 53.1 degree.

Correct. So what are the components of vector C?

 

[opus]

Thanks all! I’m about to take an exam so I’ll come back to this when I get home!