Yes, but, a few notes:
1) You can enter exercises textually:
. . . . .Given: Triangle ABC, with sides a, b, and c,
. . . . .with side a opposite the angle at point A, and
. . . . .so forth.
. . . . .Triangle ABC is right, with m(ACB) = 90°,
. . . . .m(BAC) = 30°, and m(ABC) = 60°.
. . . . .Also, the length of side b is 20 units.
. . . . .Find: the length of side a.
2) You can use what you learned back in trig and/or geometry:
. . . . .The three sides of a 30-60-90 triangle are,
. . . . .in order of length, in the ratio 1:sqrt(3):2.
. . . . .To get "20" for the "sqrt(3)" side, you'd have
. . . . .to multiply by 20/sqrt(3). Similar triangles
. . . . .have all their corresponding sides in
. . . . .proportion, so we can find the length of a,
. . . . .the shortest side, by multiplying by the same
. . . . .amount. Then:
. . . . .a = (1)(20/sqrt(3)) = 20/sqrt(3) = [20sqrt(3)]/3
This exact answer (which might be what the instructor is looking for, and his note about how to format square roots would imply that it is) evaluates to the same decimal you got.
3) You can post geometry problems to the "geometry" category.
Thank you for showing all of your work. (I reposted it as text partly to show you how it can be done, but mostly because image calls are notoriously twitchy -- I don't know why -- and later visitors might not be able to see what you're working on.)
Good work!
Eliz.
