Question

The angle of elevation of the top of the tower as observed from a point in the horizontal plane through the foot of the tower is $32{}^{\circ }.$ When the observer moves towards the tower a distance 100m, he finds the angle of elevation of the top to be $63{}^{\circ }.$ Find the height of the tower and the distance of the first position from the foot of the tower.

Answer 1

Hint: Assume that the height of the tower is h. Using $\tan \alpha ={\displaystyle \frac{AB}{AD}}$, find the length AD in terms of h.
Using $\tan \beta ={\displaystyle \frac{AB}{AC}}$ form an equation in h and hence find the value of h. Using the expression of AD in terms of h find the length AD. Hence find the distance of point C from A.

Complete step-by-step answer:

Given: AB is a tower. The angle of elevation of the tower from point C is $\beta =32{}^{\circ }$ , and the angle of elevation of the tower from point D is $\alpha =63{}^{\circ }$. DC = 100m.
To determine: The height of the tower AB and the distance of C from the foot of the tower.
Let the height of the tower be h. Hence, we have AB = h.
We have in triangle ABD, AB is the side opposite to $\alpha$ , and AD is the side adjacent to $\alpha$.
We know that $\tan \theta ={\displaystyle \frac{\text{Opposite side}}{\text{Adjacent side}}}$
Hence, we have $\tan \alpha ={\displaystyle \frac{AB}{AD}}$
Multiplying both sides by ${\displaystyle \frac{AD}{\tan \alpha }},$ we get
$AD={\displaystyle \frac{h}{\tan \alpha }}\text{ (i)}$
Also, in triangle ABC, AB is the side opposite to $\beta$ and AC is the side adjacent to $\beta$.
We know that $\tan \theta ={\displaystyle \frac{\text{Opposite side}}{\text{Adjacent side}}}$
Hence, we have $\tan \beta ={\displaystyle \frac{AB}{AC}}$
Multiplying both sides by AC, we get
$AB=AC\tan \beta$
Also, we have AC = AD +DC
Substituting the value of AD from equation (i) and substituting DC = 100, we get
$h=({\displaystyle \frac{h}{\tan \alpha }}+100)\tan \beta$
Hence, we have
$h={\displaystyle \frac{h\tan \beta }{\tan \alpha }}+100\tan \beta$
Subtracting ${\displaystyle \frac{h\tan \beta }{\tan \alpha }}$ from both sides of the equation, we get
$h-{\displaystyle \frac{h\tan \beta }{\tan \alpha }}=100\tan \beta$
Taking $h$ common from the terms on LHS, we get
$h(1-{\displaystyle \frac{\tan \beta }{\tan \alpha }})=100\tan \beta$
Dividing both sides of the equation by $1-{\displaystyle \frac{\tan \beta }{\tan \alpha }}$, we get
$h={\displaystyle \frac{100\tan \beta }{1-{\displaystyle \frac{\tan \beta }{\tan \alpha }}}}$
Substituting $\tan \beta =\tan 32{}^{\circ }=0.6248$ and $\tan \alpha =\tan 63{}^{\circ }=1.9626$, we get
$h={\displaystyle \frac{100\left(0.6248\right)}{1-{\displaystyle \frac{0.6248}{1.9626}}}}=91.67$
Hence, we have h = 91.67m
Hence the height of the tower is 91.67m.
Substituting the value of h in the equation (i), we get
$AD={\displaystyle \frac{h}{\tan \alpha }}={\displaystyle \frac{91.67}{1.9626}}=46.7$
Hence, we have AC = AD + DC = 46.7+100 = 146.7m.
Hence the distance of the first point of observation from the foot of the tower is 146.7 m.

Note: Verification:
We have AB = 91.67m, AD = 46.7m
Hence, we have ${\displaystyle \frac{AB}{AD}}={\displaystyle \frac{91.67}{46.7}}=1.9629=\tan 63{}^{\circ }$
Hence $\alpha =63{}^{\circ }$
Also, AB = 91.67m and AC = 146.7m
Hence, we have ${\displaystyle \frac{AB}{AC}}={\displaystyle \frac{91.67}{146.7}}=0.6248=\tan 32{}^{\circ }$
Hence, we have $\beta =32{}^{\circ }$
Hence our solution is verified to be correct.