Trigonometry/Angles of Elevation and Depression

Suppose you are an observer at $O$ and there is an object $Q$ , not in the same horizontal plane. Let $OP$ be a horizontal line such that $O,P,Q$ are in a vertical plane. Then if $Q$ is above $0$ , the angle $\mathrm{\angle }QOP$ is the angle of elevation of $Q$ observed from $P$ , and if $Q$ is below $O$, the angle $\mathrm{\angle }QOP$ is the angle of depression.

Often when using angle of elevation and depression we ignore the height of the person, and measure the angle from some convenient 'ground level'.

Look at the diagram above.

• $\mathrm{\angle }MLT$ is a right angle. What would you give as the translation of the labels into English?

From a point $10m$ from the base of a flag pole, its top has an angle of elevation of $50^{\circ }$ . Find the height of the pole.

[diagram]

If the height is $h$ , then $\frac{h}{10}=\tan (50^{\circ })$ . Thus $h=10\tan (50^{\circ })=11.92m$ (to two decimal places).

A flag pole is known to be $15m$ high. From what distance will its top have an angle of elevation of $50^{\circ }$ ?

[diagram]

If the distance is $d$ , then $\frac{15}{d}=\tan (50^{\circ })$ . Thus $d=\frac{15}{\tan (50^{\circ })}=12.59m$ (to two decimal places).

From the foot of a tower $20m$ high, the top of a flagpole has an angle of elevation of $30^{\circ }$ . From the top of the tower, it has an angle of depression of $25^{\circ }$ . Find the height of the flagpole and its distance from the tower.

Let the height of the flagpole be h and its distance be d. Then

(i) $\frac{h}{d}=\tan (30^{\circ })$

The top of the flagpole is below the top of the tower, since it has an angle of depression as viewed from the top of the tower. It must be $20-h$ metres lower, so

(ii) $\frac{20-h}{d}=\tan (25^{\circ })$

Adding these two equations, we find

(iii) $\frac{20}{d}=\tan (30^{\circ })+\tan (25^{\circ })$

From this (how?) we find $d=19.16m,h=11.06m$ (both to two decimal places).

From a certain spot, the top of a flagpole has an angle of elevation of $30^{\circ }$ . Move $10m$ in a straight line towards the flagpole. Now the top has an angle of elevation of $50^{\circ }$ . Find the height of the flagpole and its distance from the second point.

[diagram]

Let the height be $h$ and its distance from the second point be $x$ . Then

$\cot (50^{\circ })=\frac{x}{h};\cot (30^{\circ })=\frac{x+10}{h}$

Subtracting the first expression from the second,

$\cot (30^{\circ })-\cot (50^{\circ })=\frac{10}{h}$

$h=\frac{10}{\cot (30^{\circ })-\cot (50^{\circ })}=11.20m$

$x=h\cot (50^{\circ })=9.40m$

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