1) A plane is located at C on the diagram. There are two towers located at A and B. The distance between the towers is 7,600 feet, and the angles of elevation are given.
a) find BC, the distance from Tower 2 to the plane, to the nearest foot.
b) find CD, the height of the plane from the ground, to the nearest foot.
if you google this question, you will find the image that goes along with it. All the other sites have bad explanations and I'm very confused. Any help would be appreciated. Thank you!
I still don't understand
B = 180 − 24 =
156
ACB=
180 − 156 − 16 =
8
=
=
BC = 15052 ft
sin(B) =
sin(24) =
15052 sin(24) = x
CD = 6122 ft
If CD=h and BC=x, then
h/x = tan24°
h/(x+7600) = tan16°
or,
h = 0.4452x
h = 0.2867x + 2179.2649
0.4452x = 0.2867x + 2179.2649
BC = x = 13750
CD = h = 6121
I agree with Steve, but for finding the tan 24º that's wrong. To find the tan, it's the opposite leg divided by the adjacent leg (not the hypotenuse). So it would be h/BD=tan24º.
You could use h/x, but then it would be sin:
h/x=sin24º
since sin is the opposite leg divided by the hypotenus.
i’m praying that nunya is right, if you are wrong i will literally cry
Steve is right because it’s a right triangle
👍
so what is the answer i’m so confusedddd
Use the Law of Cosines to solve the problem. You must solve for BC first. Solve this problem in order.
A ship travels due west for 94 miles. It then travels in a northwest direction for 119 miles and ends up 173 miles from its original position. To the nearest tenth of a degree, how many degrees north of west (x) did it turn when it changed direction? Show your work.