Geometry The Law of Sines (help please)

[rachelmaddie]

I’m working on the law of sines (SSA) and the answer key is not the same solution so somehow I’m using the wrong method and I don’t know where. I’ve looked over the example a couple times and I’m very confused with this concept.

 

[MarkFL]

In the second problem, you've interchanged the sides of length 9 and 14.

 

[rachelmaddie]

So you mean it would be sin 105/14 = sin k/9?

 

[MarkFL]

So you mean it would be sin 105/14 = sin k/9?

Yes, for a proper application of the Law of Sines. You want to pair the sine of an angle with the side opposite that angle.

 

[rachelmaddie]

Yes, for a proper application of the Law of Sines. You want to pair the sine of an angle with the side opposite that angle.

Sin k = Sin-1(9sin105/14) = 8.4
m<k = 8.4
it’s not the same solution as the answer key in the textbook

 

[rachelmaddie]

Sin k = Sin-1(9sin105/14) = 8.4
m<k = 8.4
it’s not the same solution as the answer key in the textbook

I used the inverse to solve for m<k

 

[Dr.Peterson]

I don't get 8.4 degrees when I do the calculation you show, sin^-1(9*sin(105)/14), assuming that is what you meant. Can you show step by step how you did your calculation (that is, what you put into your calculator, and what it showed)?

Also, what answer does the book give?

 

[MarkFL]

Sin k = Sin-1(9sin105/14) = 8.4
m<k = 8.4
it’s not the same solution as the answer key in the textbook

I get:

[MATH]K=\arcsin\left(\frac{9}{14}\sin\left(105^{\circ}\right)\right)\approx38.38571109672363^{\circ}[/MATH]
Note: I thought I had already posted this, but somehow I didn't.

 

[rachelmaddie]

I don't get 8.4 degrees when I do the calculation you show, sin^-1(9*sin(105)/14), assuming that is what you meant. Can you show step by step how you did your calculation (that is, what you put into your calculator, and what it showed)?

Also, what answer does the book give?

I rounded to the nearest tenth.

 

[rachelmaddie]

I don't get 8.4 degrees when I do the calculation you show, sin^-1(9*sin(105)/14), assuming that is what you meant. Can you show step by step how you did your calculation (that is, what you put into your calculator, and what it showed)?

Also, what answer does the book give?

The solution in the book is 38.4

 

[rachelmaddie]

I get:

[MATH]K=\arcsin\left(\frac{9}{14}\sin\left(105^{\circ}\right)\right)\approx38.38571109672363^{\circ}[/MATH]
Note: I thought I had already posted this, but somehow I didn't.

Yes, that is the solution in the book! But I thought you’re suppose to use the inverse?

 

[Aliabla]

I'm also not getting 8.4, but It could just be my mistake. @Dr.Peterson ?
Both angle in your triangle here.

 

[rachelmaddie]

I'm also not getting 8.4, but It could just be my mistake. @Dr.Peterson ?
View attachment 13887 Both angle in your triangle here.

I’m not getting the same solution

 

[MarkFL]

Yes, that is the solution in the book! But I thought you’re suppose to use the inverse?

I prefer the notation:

[MATH]\arcsin(x)[/MATH]
Rather than:

[MATH]\sin^{-1}(x)[/MATH]
They both mean the same thing though.

 

[rachelmaddie]

I prefer the notation:

[MATH]\arcsin(x)[/MATH]
Rather than:

[MATH]\sin^{-1}(x)[/MATH]
They both mean the same thing though.

Why am I not getting the same solution

 

[rachelmaddie]

Okay, I figured it out

 

[rachelmaddie]

I prefer the notation:

[MATH]\arcsin(x)[/MATH]
Rather than:

[MATH]\sin^{-1}(x)[/MATH]
They both mean the same thing though.

This took me a while, thank you!

 

[Aliabla]

[MATH] \sin^{-1}(x)[/MATH] and [MATH]arcsine(x)[/MATH] just mean the inserve of sin it's also known as [MATH]asin[/MATH].

 

[Dr.Peterson]

Okay, I figured it out

Evidently you realized that instead of sin^-1(9*sin(105)/14) , you had somehow entered sin^-1(9*sin(105/14)).

Calculators are risky; you have to be careful how you enter an expression, and in this case look at what it displays to make sure it is what you intended.