Question

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

Answer 1

For this problem I have to come up with an example, but I'm very confused. So far I have the triangle attached as my example picture.

From there I've decided to use sin/sine:
sin45 = x/10
.71 = x/10

Can someone help me out and/or tell me if I'm even on the right track? Thanks.

Answer 2

try \[\sin(45) = \frac{ x }{ 10 }, x = 10 \times \sin(45)\]

Answer 3

How would I get to x=10×sin(45) from my original sin45 = x/10?

Answer 4

I don't know if you know this but

Answer 5

\[\sin(45) \times 10 = \frac{ x \times 10 }{ 10 }\]