Question

http://prntscr.com/amaw70
Trig question.

Answer 1

Have you written down the values of sin A, cos A, sin B and tan A in terms of a, b and c?

Answer 2

Starting with: sin A = a/c.

Answer 3

I'm sorry, but what do you mean?
@kropot72

Answer 4

The explanation here could help you to understand how the values of an angle's sine, cosine and tangent are found in a right angled triangle.
https://www.mathsisfun.com/sine-cosine-tangent.html

Answer 5

How do you add the different values in this question?

Or take them away or do anything with them?

Answer 6

So, in your question:
\[\large \sin\ A=\frac{opposite}{hypotenuse}=\frac{a}{c}\]

Answer 7

Also cos A = b/c.
Therefore:
\[\large (\sin\ A)^{2}+(\cos\ A)^{2}=\frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=\frac{a^{2}+b^{2}}{c^{2}}\]
However in any right angled triangle:
\[\large a^{2}+b^{2}=c^{2}\]
where c is the length of the hypotenuse and a and b are the lengths of the other two sides.
Can you now find whether or not the first statement in the question is true?

Answer 8

@kropot72

Not yep, but I did understand what you did until you got the the pythagorean theorem. I'm not sure how to solve if it's true with that.

Answer 9

Well it means that in the fraction:
\[\large \frac{a^{2}+b^{2}}{c^{2}}\]
the numerator and denominator are equal. So, what is the value of the fraction?

Answer 10

But how does that fraction equate to the pythagorean theorem?
@kropot72

Answer 11

The theorem means that we can replace a^2 + b^2 in the numerator with c^2, giving:
\[\large \frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=?\]

Answer 12

@kropot72

Could you show me why this theorem works?

Answer 13

The Pythagorean theorem states that in any right-angled triangle the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. Therefore in the diagram in the question we can write:
\[\large a^{2}+b^{2}=c^{2}\]
Do you understand this?

Answer 14

Yes, I understand the Pythagorean theorem.

How it fits into the picture is confusing me a bit because how is a^2+b^2/c^2=a^2+b^2=c^2?

@kropot72

Answer 15

@knov

Mind continuing since he isn't here?

Answer 16

sure