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

Answer 1

(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.