Question

the angle of elevation of the top of a tower as observed from a point in a horizontal line through the foot of the tower is 30 degree when the observer moves towards the tower a distance of 100 meters he finds the angle of elevation to the top to be 60 degree find the height of the tower and distance of the firt position from the tower

Answer 1

Given: The angle of elevation in the first instance is 30° while in the second it is 60°.

To find: The height of the tower and the distance between the first point of elevation and the tower.

Answer:

(Diagram for reference attached below.)

In Δ ABC,

$\tt tan {30}^{\circ}\ =\ \dfrac{AB}{BC}\\\\\\\bigg[tan\ {30}^{\circ}\ =\ \dfrac{1}{\sqrt{3}}\bigg]\\\\\\\dfrac{1}{\sqrt{3}}\ =\ \dfrac{AB}{BD\ +\ DC}\\\\\\\ \dfrac{1}{\sqrt{3}}\ =\ \dfrac{AB}{BD\ +\ 100}\\\\\\AB\ =\ \dfrac{BD\ +\ 100}{\sqrt{3}}$

In Δ ABD,

$\tt tan\ {60}^{\circ}\ =\ \dfrac{AB}{BD}\\\\\\\bigg[tan\ {60}^{\circ}\ =\ \sqrt{3}\bigg]\\\\\\\sqrt{3}\ =\ \dfrac{AB}{BD}\\\\\\AB\ =\ \sqrt{3}BD\\\\\\\dfrac{BD\ +\ 100}{\sqrt{3}}\ =\ \sqrt{3}BD\\\\\\BD\ +\ 100\ =\ 3BD\\\\\\100\ =\ 3BD\ -\ BD\\\\\\100\ =\ 2BD\\\\\\\dfrac{100}{2}\ =\ BD\\\\\\50\ m\ =\ BD$

Therefore, the distance between the tower and the first point of elevation is BD + DC = 50 + 100 = 150 m.

As we saw earlier,

$\tt AB\ =\ \dfrac{BD\ +\ 100}{\sqrt{3}}$

Using the value of BD that we obtained,

$\tt AB\ =\ \dfrac{50\ +\ 100}{\sqrt{3}}\\\\\\AB\ =\ \dfrac{150}{\sqrt{3}}\\\\\\AB\ =\ 50\sqrt{3}\ m$

Therefore, the height of the tower is 50√3 m.

Answer 2

Answer:

Given: The angle of elevation in the first instance is 30° while in the second it is 60°.

To find: The height of the tower and the distance between the first point of elevation and the tower.

Answer:

(Diagram for reference attached below.)

In Δ ABC,

\begin{gathered}\tt tan {30}^{\circ}\ =\ \dfrac{AB}{BC}\\\\\\\bigg[tan\ {30}^{\circ}\ =\ \dfrac{1}{\sqrt{3}}\bigg]\\\\\\\dfrac{1}{\sqrt{3}}\ =\ \dfrac{AB}{BD\ +\ DC}\\\\\\\ \dfrac{1}{\sqrt{3}}\ =\ \dfrac{AB}{BD\ +\ 100}\\\\\\AB\ =\ \dfrac{BD\ +\ 100}{\sqrt{3}}\end{gathered}tan30∘ = BCAB[tan 30∘ = 31]31 = BD + DCAB 31 = BD + 100ABAB = 3BD + 100

In Δ ABD,

\begin{gathered}\tt tan\ {60}^{\circ}\ =\ \dfrac{AB}{BD}\\\\\\\bigg[tan\ {60}^{\circ}\ =\ \sqrt{3}\bigg]\\\\\\\sqrt{3}\ =\ \dfrac{AB}{BD}\\\\\\AB\ =\ \sqrt{3}BD\\\\\\\dfrac{BD\ +\ 100}{\sqrt{3}}\ =\ \sqrt{3}BD\\\\\\BD\ +\ 100\ =\ 3BD\\\\\\100\ =\ 3BD\ -\ BD\\\\\\100\ =\ 2BD\\\\\\\dfrac{100}{2}\ =\ BD\\\\\\50\ m\ =\ BD\end{gathered}tan 60∘ = BDAB[tan 60∘ = 3]3 = BDABAB = 3BD3BD + 100 = 3BDBD + 100 = 3BD100 = 3BD − BD100 = 2BD2100 = BD50 m = BD

Therefore, the distance between the tower and the first point of elevation is BD + DC = 50 + 100 = 150 m.

As we saw earlier,

\tt AB\ =\ \dfrac{BD\ +\ 100}{\sqrt{3}}AB = 3BD + 100

Using the value of BD that we obtained,

\begin{gathered}\tt AB\ =\ \dfrac{50\ +\ 100}{\sqrt{3}}\\\\\\AB\ =\ \dfrac{150}{\sqrt{3}}\\\\\\AB\ =\ 50\sqrt{3}\ m\end{gathered}AB = 350 + 100AB = 3150AB = 503 m

Therefore, the height of the tower is 50√3 m.