# Trigonometry/Angles of Elevation and Depression

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

 Exercise 1: Opposite, Hypotenuse, Adjacent Look at the diagram above. ${\displaystyle \angle MLT}$ is a right angle. What would you give as the translation of the labels into English?
 Example 1: A Flagpole From a point ${\displaystyle 10m}$ from the base of a flag pole, its top has an angle of elevation of ${\displaystyle 50^{\circ }}$ . Find the height of the pole. [diagram] If the height is ${\displaystyle h}$ , then ${\displaystyle {\frac {h}{10}}=\tan(50^{\circ })}$ . Thus ${\displaystyle h=10\tan(50^{\circ })=11.92m}$ (to two decimal places).
 Example 2: A ${\displaystyle 15m}$ High Flagpole A flag pole is known to be ${\displaystyle 15m}$ high. From what distance will its top have an angle of elevation of ${\displaystyle 50^{\circ }}$ ? [diagram] If the distance is ${\displaystyle d}$ , then ${\displaystyle {\frac {15}{d}}=\tan(50^{\circ })}$ . Thus ${\displaystyle d={\frac {15}{\tan(50^{\circ })}}=12.59m}$ (to two decimal places).
 Example 3: A ${\displaystyle 20m}$ Tower From the foot of a tower ${\displaystyle 20m}$ high, the top of a flagpole has an angle of elevation of ${\displaystyle 30^{\circ }}$ . From the top of the tower, it has an angle of depression of ${\displaystyle 25^{\circ }}$ . 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) ${\displaystyle {\frac {h}{d}}=\tan(30^{\circ })}$ 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 ${\displaystyle 20-h}$ metres lower, so (ii) ${\displaystyle {\frac {20-h}{d}}=\tan(25^{\circ })}$ Adding these two equations, we find (iii) ${\displaystyle {\frac {20}{d}}=\tan(30^{\circ })+\tan(25^{\circ })}$ From this (how?) we find ${\displaystyle d=19.16m\ ,\ h=11.06m}$ (both to two decimal places).
 Example 4: Yet another Flagpole From a certain spot, the top of a flagpole has an angle of elevation of ${\displaystyle 30^{\circ }}$ . Move ${\displaystyle 10m}$ in a straight line towards the flagpole. Now the top has an angle of elevation of ${\displaystyle 50^{\circ }}$ . Find the height of the flagpole and its distance from the second point. [diagram] Let the height be ${\displaystyle h}$ and its distance from the second point be ${\displaystyle x}$ . Then ${\displaystyle \cot(50^{\circ })={\frac {x}{h}}\ ;\ \cot(30^{\circ })={\frac {x+10}{h}}}$ Subtracting the first expression from the second, ${\displaystyle \cot(30^{\circ })-\cot(50^{\circ })={\frac {10}{h}}}$ ${\displaystyle h={\frac {10}{\cot(30^{\circ })-\cot(50^{\circ })}}=11.20m}$ ${\displaystyle x=h\cot(50^{\circ })=9.40m}$