+0

# Applications to Trignometry

+5
288
6
+302

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)

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.

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.

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.

I need the answers to all these questions. I would like you to show the working with diagram if possible.

Thanks,

SARAH

SARAHann  Mar 11, 2017

#5
+79735
+10

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....

CPhill  Mar 12, 2017
Sort:

#1
+79735
+10

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

CPhill  Mar 11, 2017
#2
+79735
+10

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

CPhill  Mar 11, 2017
#3
+79735
+10

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

CPhill  Mar 12, 2017
#4
+79735
+10

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)

CPhill  Mar 12, 2017
#5
+79735
+10

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....

CPhill  Mar 12, 2017
#6
+302
+5

Thank you CPhill so much for answering my Questions! i can't thank you enough! I am so glad for this amazing website that you guys have created and thank you all so much! ;_)

SARAHann  Mar 12, 2017

### 14 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details