(i) Distance RT
Let x be the distance RT.
Using the Pythagorean Theorem, we have:
x^2 = (20m)^2 + (3m)^2
x^2 = 400m^2 + 9m^2
x^2 = 409m^2
x = √409m = 20.2m
Therefore, the distance RT is 20.2m.
(ii) Distance of the foot of the taller pole from P
Let y be the distance of the foot of the taller pole from P.
Using the Law of Cosines, we have:
y^2 = (20m)^2 + (7m)^2 - 2(20m)(7m)cos30°
y^2 = 400m^2 + 49m^2 - 280m^2cos30°
y^2 = 169m^2 - 168m^2cos30°
y^2 = 169m^2 - 84m^2
y^2 = 85m^2
y = √85m = 9.2m
Therefore, the distance of the foot of the taller pole from P is 9.2m, correct to 3s.f.
(iii) Angle of elevation of T from P
Let θ be the angle of elevation of T from P.
Using the Law of Sines, we have:
sinθ = (7m)/y
sinθ = (7m)/9.2m
sinθ = 0.76
θ = sin^-1(0.76)
θ = 57.2°
Therefore, the angle of elevation of T from P is 57.2°, correct to 1d.p.
The feet of two vertical poles of heights 3metres and 7metres are in line with a point P on the ground, the smaller pole being between the taller pole and P at a distance 20metres from P. The angle of elevation of the top (T) of the taller pole from the top (R) of the smaller pole is 30◦. Calculate the: (i) distance RT (ii) distance of the foot of the taller pole from P, correct to 3s.fs. (iii) angle of elevation of T from P, correct to 1d.p. Please with full and well detailed and better solution that is done by math teachers that know what they are doing
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