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
STOP IT!
(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.