Question

Really need help;
(Posting in comments)

Answer 1

In the right triangle ABC,\[m<B = 90°\] and \[\cos A = \frac{ 11 }{ 17 }\] What are m<A and m<C?

Answer 2

@iGreen. @Legends. Can either of you help?

Answer 3

o no, i can't help, im not good at trig, sorry

Answer 4

No, I have not learned this yet. @Luigi0210

Answer 5

Okay, so you have a triangle like so right?:
And they give you cos A =11/17. Meaning that the angle A, you have sides 11 and 17

Answer 6

I just tried visualizing it using GeoGebra with the lengths they gave me and I got the answers. It took me a minute but I understand what it's asking now.
m<A = 50 (Roughly) and m<C = 40 (Roughly)

Answer 7

Oh..that's nice, but you might want to learn how to do it by hand.

Answer 8

If @Luigi0210 wants to explain it, I'm more than willing to learn to do it.

Answer 9

So we'll just make A a random place on the triangle:

Since \(\Large \cos = \frac{adi}{hyp} \)

Answer 10

Now do I use cos-1 to find the angle of m<A?

Answer 11

I got 49.684 on cos-1, so I'm assuming that's how to get m<A. What about m<C?

Answer 12

Yea, visualizations just help so you don't make mistakes.

So for A: \(\Large A= cos^{-1} (\frac{11}{17})\)
B: \(\Large B = sin^{-1} \frac{11}{17} \)

Answer 13

Whoops, that should be C not B

Answer 14

Because cos-1 for A is the same as sin-1 for B?

Answer 15

But since a triangle is 180, and you have 90 and ~50, you can add those two sides and it'll give you the missing side.

\(\Large 180=90+49.684 + C\)

Just solve like an equation ~

Answer 16

Missing side will be about ~40

Answer 17

So that would give\[180 = ~140 + C\] And, to find C you have to get it by itself, so: \[~40 = C\]

Answer 18

Thanks @Luigi0210 and @iGreen!

Answer 19

Yup yup

It'll get more exciting when you get into Law of Sines and Cosines :D