Question

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.
c.)Find the sine, cosine, and tagent of the 30* angle in the lot. Write your answers as decimals rounded to four decimal places.

Answer 1

PLEASE HELP!!!

Answer 2

sin 30 = 41/h
1/2 = 41/h
h = 2*41 (its also the answer for part b)

a. using Pythagoras theorem.
\[a^2 +o^2 = h^2\]
we have o = 41
h = 82
now find a

c. sin 30 = Opposite/ hypotenuse
cos 30 = adjacent /hypotenuse
tan 30 = opposite/adjacent
opposite given in question
hypotenuse found in part b
adjacent found in part a

Answer 3

can you help me further I don't understand or have a calculator lol

Answer 4

sin 30 = 41/h
1/2 = 41/h
h = 2*41 (its also the answer for part b)
h = 82 feet

a. using Pythagoras theorem.
\[a^2+o^2=h^2\]
we have o = 41
h = 82
\[a^2+41^2=82^2\]
\[a^2+1681=6724\]
\[a^2=6724 -1681\]
\[a^2=5043\]
\[a=\sqrt{5043}\]

c. sin 30 = Opposite/ hypotenuse
\[\sin 30^0 = 41/82= 1/2\]
cos 30 = adjacent /hypotenuse
\[\cos 30^0 =\sqrt{5043}/82 = 71.02/82\]
tan 30 = opposite/adjacent
\[\tan 30^0 = 41/71.02\]