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
(i) 4/RT = sin30° so RT = 8
(ii) 4√3 + 20 = 26.93
(iii) tanθ = 7/26.93 so θ = 14.57°
(i) Distance RT:
Using the Pythagorean Theorem, we can calculate the distance RT as follows:
RT = √(20² + (7-3)²)
RT = √(400 + 16)
RT = √416
RT = 20.4 m
(ii) Distance of the foot of the taller pole from P:
Using the Pythagorean Theorem, we can calculate the distance of the foot of the taller pole from P as follows:
PT = √(20² + 7²)
PT = √(400 + 49)
PT = √449
PT = 21.1 m
Correct to 3s.f.s:
PT = 21.1 m
PT = 21.1 m
(iii) Angle of elevation of T from P:
Using the tangent ratio, we can calculate the angle of elevation of T from P as follows:
tan θ = Opposite/Adjacent
tan θ = 7/20
θ = tan⁻¹(7/20)
θ = 24.3°
