Question

I just need help to start off like an equation then I can figure it out.
A building lot in a city is shaped as a 30° -60° -90° triangle. The side opposite the 30° angle measures 41 feet.

a. Find the length of the side of the lot opposite the 60° angle.
b. Find the length of the hypotenuse of the triangular lot.

Answer 1

\[\sin 30^{o} = 0.5 = x/h\]\[h = x / \sin 30^{o} = 41 / 0.5 = 82\]So, that's the length of the hypotenuse. For the other side, you can use either trig or the Pythagorean Theorem. Since we just did trig, let's use instead:\[x ^{2} + y ^{2} = h ^{2}\]\[y = \sqrt{h ^{2} - x ^{2}} = \sqrt{82^{2} - 41^{2}}\]

Answer 2

It's been a while this should help me get started thanks.

Answer 3

uw!

Answer 4

Find the sine, cosine, and tangent of the 30° angle in the lot. Write your answers as decimals rounded to four decimal places. do i just change where you put sin to the cos,tan

Answer 5

sin 30 = x/h = opposite over hypotenuse
cos 30 = y/h = adjacent over hypotenuse
tan 30 = x/y = opposite over adjacent.

Answer 6

There's a little bit more for you, @mym2e . Hope that helps.

Answer 7

And you now have the measures of "x", "y", and "h" from my earlier replies.

Answer 8

You also can use this "standard" triangle to get the trig ratios:

Answer 9

Here, tan 30 = (1/2) / ([sqrt(3)]/2) = [sqrt(3)]/3

Answer 10

It's the same as:

tan 30 = 41 / sqrt(82^2 - 41^2)

from the other diagram. It doesn't matter what side lengths we are talking about. The trig ratio will always be the same. Whether a large or small triangle, sin 30 is always sin 30, etc.