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 solution
(i) Distance RT
Let RT = x
Using the Pythagorean Theorem,
x^2 = (7m)^2 + (20m)^2
x^2 = 4900m^2 + 400m^2
x^2 = 5300m^2
x = √5300m
x = 73.02m
Distance RT = 73.02m
(ii) Distance of the foot of the taller pole from P
Let TP = y
Using the Pythagorean Theorem,
y^2 = (7m)^2 + (x)^2
y^2 = 4900m^2 + (73.02m)^2
y^2 = 4900m^2 + 5369.0604m^2
y^2 = 10269.0604m^2
y = √10269.0604m
y = 101.25m
Distance of the foot of the taller pole from P = 101.25m, correct to 3s.fs.
(iii) Angle of elevation of T from P
Let angle PTR = θ
Using the Law of Sines,
sinθ = (7m)/(101.25m)
sinθ = 0.0693
θ = sin^-1(0.0693)
θ = 4.06°
Angle of elevation of T from P = 4.1°, correct to 1d.p.