Question

A point H is 20m away from the foot of a tower on the same horizontal ground. From the point H, the angle of elevation of the point (P) on the tower and the top(T) of the tower are 30° and 50° respectively.
( a) draw a diagram to illustrate the information above.
(b) calculate correct to 3 s.f,
( I) /PT/
(ii) the distance between H and the too of the tower.
(III) the position of H if the angle of depression of H from the too of the tower is to be 40°

Answer 1

Answer:

a. See Attachment 1

b. $PT = 12.3\ m$

c. $HT = 31.1\ m$

d. $OH = 28.4\ m$

Step-by-step explanation:

Calculating PT

To calculate PT, we need to get distance OT and OP

Calculating OT;

We have to consider angle 50, distance OH and distance OT

The relationship between these parameters is;

$tan50 = \frac{OT}{20}$

Multiply both sides by 20

$20 * tan50 = \frac{OT}{20} * 20$

$20 * tan50 = OT$

$20 * 1.1918 = OT$

$23.836 = OT$

$OT = 23.836$

Calculating OP;

We have to consider angle 30, distance OH and distance OP

The relationship between these parameters is;

$tan30 = \frac{OP}{20}$

Multiply both sides by 20

$20 * tan30 = \frac{OP}{20} * 20$

$20 * tan30 = OP$

$20 * 0.5774= OP$

$11.548 = OP$

$OP = 11.548$

$PT = OT - OP$

$PT = 23.836 - 11.548$

$PT = 12.288$

$PT = 12.3\ m$ (Approximated)

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Calculating the distance between H and the top of the tower

This is represented by HT

HT can be calculated using Pythagoras theorem

$HT^2 = OT^2 + OH^2$

Substitute 20 for OH and $OT = 23.836$

$HT^2 = 20^2 + 23.836^2$

$HT^2 = 400 + 568.154896$

$HT^2 = 968.154896$

Take Square Root of both sides

$HT = \sqrt{968.154896}$

$HT = 31.1\ m$ (Approximated)

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Calculating the position of H

This is represented by OH

See Attachment 2

We have to consider angle 50, distance OH and distance OT

The relationship between these parameters is;

$tan50 = \frac{OH}{OT}$

Multiply both sides by OT

$OT * tan50 = \frac{OH}{OT} * OT$

$OT * tan50 = {OH$

$OT * 1.1918 = OH$

Substitute $OT = 23.836$

$23.836 * 1.1918 = OH$

$28.4= OH$

$OH = 28.4\ m$ (Approximated)