Finding sides AC and BC in a 60-90-30 right triangle

[xyz_1965]

Right triangle ABC is given with angles A, B, and C, where angle C is 90 degrees. Angle A is 60° and side AB = 12 cm. Find sides AC and BC.

Here is the set up.

To find AC:

cos (60°) = AC/12

To find BC:

sin (60°) = BC/12

Is this correct?

 

[HOI]

Yes. With the usual notation, AB is the side opposite C, the right angle, so is the hypotenuse. Since you are given angle A, AC is the "near side" and BC is the "opposite side". cos(A) is "near side over hypotenuse" and sin(A) is "opposite side over hypotenuse".

Hopefully, you will soon have enough confidence in yourself that you won't need to ask questions like these!

 

[xyz_1965]

Yes. With the usual notation, AB is the side opposite C, the right angle, so is the hypotenuse. Since you are given angle A, AC is the "near side" and BC is the "opposite side". cos(A) is "near side over hypotenuse" and sin(A) is "opposite side over hypotenuse".

Hopefully, you will soon have enough confidence in yourself that you won't need to ask questions like these!

I hope to get there soon. I found a few questions that I am stuck with in terms of a geometric figure. I will post each question later. I simply need help setting it up. If you can provide me with a picture, a visual of the situation, that would be so cool and helpful.