Answer:
a. See Attachment 1
b. ![PT = 12.3\ m](https://tex.z-dn.net/?f=PT%20%3D%2012.3%5C%20m)
c. ![HT = 31.1\ m](https://tex.z-dn.net/?f=HT%20%3D%2031.1%5C%20m)
d. ![OH = 28.4\ m](https://tex.z-dn.net/?f=OH%20%3D%2028.4%5C%20m)
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}](https://tex.z-dn.net/?f=tan50%20%3D%20%5Cfrac%7BOT%7D%7B20%7D)
Multiply both sides by 20
![20 * tan50 = \frac{OT}{20} * 20](https://tex.z-dn.net/?f=20%20%2A%20tan50%20%3D%20%5Cfrac%7BOT%7D%7B20%7D%20%2A%2020)
![20 * tan50 = OT](https://tex.z-dn.net/?f=20%20%2A%20tan50%20%3D%20OT)
![20 * 1.1918 = OT](https://tex.z-dn.net/?f=20%20%2A%201.1918%20%3D%20OT)
![23.836 = OT](https://tex.z-dn.net/?f=23.836%20%20%3D%20OT)
![OT = 23.836](https://tex.z-dn.net/?f=OT%20%3D%2023.836)
Calculating OP;
We have to consider angle 30, distance OH and distance OP
The relationship between these parameters is;
![tan30 = \frac{OP}{20}](https://tex.z-dn.net/?f=tan30%20%3D%20%5Cfrac%7BOP%7D%7B20%7D)
Multiply both sides by 20
![20 * tan30 = \frac{OP}{20} * 20](https://tex.z-dn.net/?f=20%20%2A%20tan30%20%3D%20%5Cfrac%7BOP%7D%7B20%7D%20%2A%2020)
![20 * tan30 = OP](https://tex.z-dn.net/?f=20%20%2A%20tan30%20%3D%20OP)
![20 * 0.5774= OP](https://tex.z-dn.net/?f=20%20%2A%200.5774%3D%20OP)
![11.548 = OP](https://tex.z-dn.net/?f=11.548%20%3D%20OP)
![OP = 11.548](https://tex.z-dn.net/?f=OP%20%3D%2011.548)
![PT = OT - OP](https://tex.z-dn.net/?f=PT%20%3D%20OT%20-%20OP)
![PT = 23.836 - 11.548](https://tex.z-dn.net/?f=PT%20%3D%2023.836%20-%2011.548)
![PT = 12.288](https://tex.z-dn.net/?f=PT%20%3D%2012.288)
(Approximated)
--------------------------------------------------------
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](https://tex.z-dn.net/?f=HT%5E2%20%3D%20OT%5E2%20%2B%20OH%5E2)
Substitute 20 for OH and ![OT = 23.836](https://tex.z-dn.net/?f=OT%20%3D%2023.836)
![HT^2 = 20^2 + 23.836^2](https://tex.z-dn.net/?f=HT%5E2%20%3D%2020%5E2%20%2B%2023.836%5E2)
![HT^2 = 400 + 568.154896](https://tex.z-dn.net/?f=HT%5E2%20%3D%20400%20%2B%20568.154896)
![HT^2 = 968.154896](https://tex.z-dn.net/?f=HT%5E2%20%3D%20968.154896)
Take Square Root of both sides
![HT = \sqrt{968.154896}](https://tex.z-dn.net/?f=HT%20%3D%20%5Csqrt%7B968.154896%7D)
<em>(Approximated)</em>
--------------------------------------------------------
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}](https://tex.z-dn.net/?f=tan50%20%3D%20%5Cfrac%7BOH%7D%7BOT%7D)
Multiply both sides by OT
![OT * tan50 = \frac{OH}{OT} * OT](https://tex.z-dn.net/?f=OT%20%2A%20tan50%20%3D%20%5Cfrac%7BOH%7D%7BOT%7D%20%2A%20OT)
![OT * tan50 = {OH](https://tex.z-dn.net/?f=OT%20%2A%20tan50%20%3D%20%7BOH)
![OT * 1.1918 = OH](https://tex.z-dn.net/?f=OT%20%2A%201.1918%20%3D%20OH)
Substitute ![OT = 23.836](https://tex.z-dn.net/?f=OT%20%3D%2023.836)
![23.836 * 1.1918 = OH](https://tex.z-dn.net/?f=23.836%20%2A%201.1918%20%3D%20OH)
![28.4= OH](https://tex.z-dn.net/?f=28.4%3D%20OH)
<em> (Approximated)</em>