Question

Two lamp posts are of equal height. A boy standing mid-way between then observes the elevation of the top of either post to be 30∘. After walking 15 m towards one of them, he observes the elevation of it stop to be 45∘. Find the heights of the posts and the distance between them.​(Ans:20.49m,70.98m)

Answer 1

The height of the lamp posts is 20.49 m and the distance between them is 70.98 m.

Explanation:

Draw a diagram using the given information (see attachment).

• The lamp post is at point C (the top of the lamp post is at point D).
• The height of the lamp post is labelled "h".
• The boy is at point A.
• After walking 15 m towards one of the lamp posts, the boy is at point B.
• The distance between both lamp posts is 2x.

`\boxed{\begin{minipage}{7 cm}\underline{Tan trigonometric ratio} \\\\$\sf \tan(\theta)=(O)/(A)$\\\\where:\\ \phantom{ww}$\bullet$ $\theta$ is the angle. \\ \phantom{ww}$\bullet$ $\sf O$ is the side opposite the angle. \\\phantom{ww}$\bullet$ $\sf A$ is the side adjacent the angle. \\\end{minipage}}`

Use the tan trigonometric ratio to create two equations for h.

Triangle ACD:

`\implies \tan 30^(\circ)=(h)/(x)`

`\implies h=x\tan 30^(\circ)`

`\implies h=(√(3))/(3)x`

Triangle BCD:

`\implies \tan 45^(\circ)=(h)/(x-15)`

`\implies h=(x-15)\tan 45^(\circ)`

`\implies h=(x-15)\cdot 1`

`\implies h=x-15`

Substitute the second equation into the first equation and solve for x:

`\implies x-15=(√(3))/(3)x`

`\implies x-(√(3))/(3)x=15`

`\implies (3-√(3))/(3)x=15`

`\implies (3-√(3))x=45`

`\implies x=(45)/(3-√(3))`

`\implies x=35.4903810...`

To find h, substitute the found value of x into the second equation and solve for h:

`\implies h=x-15`

`\implies h=35.4903810...-15`

`\implies h=20.4903810...`

`\implies h=20.49\; \sf m\;\;(2\;d.p.)`

As the distance between the two lamp posts is 2x:

`\implies 2x=2 \cdot 35.4903810...`

`\implies 2x=70.9807621...`

`\implies 2x=70.98\; \sf m\;\;(2\;d.p.)`

Therefore:

• The height of the lamp posts is 20.49 m.
• The distance between them is 70.98 m.