Applications to Trignometry

4. The angle of elevation of the top B of a tower AB from a point X on the ground is 60°. at a point Y 40m vertically above X, the angle of elevation of the top is 45°. find the height of the tower AB and the distance XB.

Let D be the horizontal distance from X to the tower base

Let H be the tower height

And we have :

tan 60  = H/D  →  D = H / tan 60

And we know that....the vertical difference between the tower's height and Y = (H - 40)...so...

tan 45  = (H - 40) / D       sub for D

tan 45  = (H - 40) / [H/tan 60 ]

tan 45  = tan 60 *(H - 40) / H

H *tan 45  = tan 60 (H - 40)

H * tan 45  = H* tan 60  = 40 tan 60

H = H * sqrt(3) - 40sqrt(3)

40sqrt3 = H [ sqrt(3) - 1]

[40sqrt(3)] /[ sqrt(3) -1]  = H  ≈ 94.64 m

And XB ≈ H csc 60 ≈ 94.64 csc 60 ≈ 109.28 m

Here's a pic....

1. From the top of a tower of height 50m, the angles of depression of the top and bottom of a pole are 30° and 45° respectively. Find 

a) how far the pole is from the bottom of the tower

b) the height of the pole (use root3 =1.732)

a) The distance, D, is given by :

D = 50 / tan(45)  = 50m

b) The height of the pole, H, can be found by

H =  50 - 50tan(30)   =  50 - 50/sqrt(3)   =  50 - 50/1.732  ≈ 21.13 m

Here's a pic for 1

AB is the pole = 21.13m, DC is the tower = 50m....BD = 50m..angle ECA = 30....angle ECB  = 45

2. The horizontal distanc between two towers is 60m. the angle of elevation of the top of the taller tower as seen from the top of the shorter one is 30°.If the height of the taller tower is 150m, then find the height of the shorter tower.

The height of the shorter tower  =

150 - 60tan30  =  150 - 60sqrt(3)  ≈  115.36 m

Here's a pic....BA   = shorter tower = 115.35m  .....CD = taller tower = 150m....angle CAF = 30.....AF = 60m

3. From the top of the tower the angle of depression of an object on the horizontal ground is found to be 60°. on decending 20m vertically downwards from the top of the tower, the angle of depression of the object is found to be 30°. find the height ofthe tower.

Let O   be the object's distance from the tower's base  and H  be the tower height....and we have that

tan (60)  = H/O   →  O  = H/tan (60) = H/sqrt(3)

And

tan (30)  = (H - 20) / O              sub  H/ tan(60) for O

tan (30)  = (H - 20) / [H / tan(60)]

tan (30)  = tan(60) ( H - 20) / H

H tan (30)  = tan(60) (H - 20)

Htan(30) = Htan(60) - 20tan(60)

20  tan(60)  = H tan(60) - H tan(30)

20tan(60)  = H [tan (60) - tan (30)]

[20tan(60) ]/ [ tan(60) - tan (30) ]  = H  = 30 m

Here's a pic  AB  = tower height  = 30m   ....OA  = H/sqrt(3)  = 30/sqrt(3)