A tower stand verticle from the ground from a point on ground which is 100m away from the foot of tower and angle of elevation from the top of lower is found to be 60. Find the height of the tower.
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A tower stand verticle from the ground from a point on ground which is 100m away from the foot of tower and angle of elevation from the top of lower is found to be 60. Find the height of the tower.
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Answer:
Step-by-step explanation:
In right triangle ABC, the angle at B is 60 degrees and AC = 100 meters. Thus, BC = 100 cot(60 degrees) = 100 sqrt(3)/3 or 100 sqrt(3)/3 meters. AB = BC + AC = 100 sqrt(3)/3 + 100 meters = 100(1 + sqrt(3))/3 meters. Thus, the height of the tower is 100 meters. (Note that we could have also used the formula tan(60 degrees) = BC/AC = height of tower / 100 meters, which gives us the height of the tower = 100 tan(60 degrees) = 100 sqrt(3) = 100(1 + sqrt(3))/3 meters.) [asy]
import TrigMacros;
size(200);
pair A, B, C, F;
A = (0,100);
B = (100,0);
C = IP(arc(A,100,0,60),B--A);
F = foot(A,C,B);
draw(A--B--C--cycle);
draw(A--F);
label("A", A, N);
label("B", B, S);
label("C", C, E);
label("F", F, N);
label("100",(A + C)/2, NW);
label("100",(B + F)/2, W);
label("60
∘
", (A + B)/2, S);
Verified answer
[tex]\bfAnswer:[/tex]
To find the height of the tower, you can use trigonometry. The angle of elevation is 60 degrees, and the horizontal distance from the point on the ground to the foot of the tower is 100 meters. You can use the tangent function (tan) to find the height (h) of the tower:
tan(60 degrees) = h / 100 meters
Now, you can solve for h:
h = 100 meters * tan(60 degrees)
h ≈ 173.2 meters
So, the height of the tower is approximately 173.2 meters.