Answer:
The height of the tower is 130.5m
Step-by-step explanation:
In the question, we are given the following values:
The angles of elevation to the top of the tower from P = 50°
The angles of elevation to the top of the tower from Q = 45°
The angles of elevation to the top of the tower from P = 50°
Hence,cot P = 1/ tan(50°)
The angles of elevation to the top of the tower from Q = 45°
Hence, tan 45° = 1
In the question,we are told Points P and Q lie 240 m apart in line with and on opposite sides of a communications tower.
Therefore,
PQ = height of the tower( tan Q + 1/tan P)
240m = height of the tower( tan 45° + 1/ tan 50°)
240m = h(1 + 1/tan 50°)
h = (240 m)/(1 + 1/tan (50°))
h = 130.49863962 meters
Therefore, the height of the tower to the nearest tenth of a meter is 130.5 meters(m)
Height of the tower is 130.5 meters.
Given in the question,
Let the height of the tower = h m
And the distance of P from the tower = x m
From ΔPSR,
[tex]\text{tan}(\angle P)=\frac{RS}{PS}[/tex]
[tex]\text{tan}(50^\circ)=\frac{h}{240-x}[/tex]
240 - x = h.cot(50°)
x = 240 - hcot(50°) ------(1)
From ΔQSR,
[tex]\text{tan}(\angle Q)=\frac{RS}{SQ}[/tex]
[tex]\text{tan}(45^\circ)=\frac{h}{x}[/tex]
x = h.cot(45°) ------(2)
Equate the values of x from equations (1) and (2),
240 - h.cot(50°) = h.cot(45°)
h[cot(50°)+cot(45°)]=240
h = [tex]\frac{240}{1.8391}[/tex]
h = 130.498
h ≈ 130.5 m
Therefore, height of the tower is 130.5 meters.
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