Question

How could I find the two missing side lengths of a right triangle if I only know one side length and one angle measure (other than the 90 degree angle)?

Answer 1

You can always get the answer using either sin, cos or tan depending on which side you know and how its positioned relative to the angle you know.

Answer 2

Using a trig ratio (sin, cos, tan) I'm assuming I could, but I'm confused on how to actually solve and do it. Using this example that I just made up, could someone help me understand it?

Answer 3

Oh haha, guess we had the same thought.

Answer 4

Lets find BC. Remember\[\cos(x)=\frac{ adjacent }{ hypotenuse }\]

Answer 5

So
\[\cos(45)=\frac{ BC }{ 10 }\]

Answer 6

Pretend like you are sitting on the angle you know. BC is the "adjacent side" AB is the "opposite side" 10 is the "hypotenuse"

Answer 7

Can you solve that equation for BC?

Answer 8

Are you doing trig functions in this class or whatever? There's actually a non-trig way to do this one when the angles are 45 - 45 - 90. I assumed you were doing trig.

Answer 9

Yeah I've seen both 454590 and 306090 triangles, but I don't think that they want me to use that for these type of problems. But one second, I'm gonna jot down everything you've said real quick so I can try and solve it.

Answer 10

Okay can you give me a hint on what to do to solve the cos45 = BC/10 because I'm stuck.

Answer 11

Would I figure out the cos first or would I divide/multiply or what? I used to know this but nothing's clicking right now haha.

Answer 12

1st multiply both sides by 10.
You'll get 10cos45 = BC
Then use a calator..
Or if you know cos(45) = sqrt(2) / 2 do it that way

Answer 13

\[10\cos(45)=10\frac{ \sqrt{2} }{ 2 }=5\sqrt{2}\]

Answer 14

Gotta go. Good luck!

Answer 15

Alright! Thanks for all your help!

Answer 16

Okay I've taken everything BangkokGarrett told me to do:
Multiply both sides by 10
10cos45 = BC

and ended up with:
7.07

Can someone tell me if I've done this right?

Answer 17

Here's my entire work-through of the problem, can someone let me know if I've done it right?

Answer 18

@BangkokGarrett Hey if you're busy ignore this, but I'm wondering if you can look at my work and tell me if I've done it right, somewhat right, or right at all lol.

Answer 19

5root(2) and 7.07 are the same thing. Yes, you're ok.

Answer 20

Sweet! How would I find the 5root(2) answer instead of the 7.07? Would I inverse my 7.07 answer or what?

Answer 21

There's no way you could look at that 7.07 your calculator spits out and notice "oh, that's 5 times the square root of 2" unless you're a total math prodigy. The only way you could have the root 2 in your answer if you recognize that cos(45) = root 2 / 2.
Many math teachers have students memorize the sins and coss of common angles ... like 45 degrees

Answer 22

Oh I gotcha. I just know that the lessons I was reading talked about the two different answer types, but I suppose it doesnt *really* matter, both are correct I'm assuming.