Angle of Elevation and Depression: Class 10 Maths Guide

Stand at the foot of a tall pole, a flag­pole or a minaret and tip your head back to look at its top. Your line of sight now makes an angle with the hor­i­zon­tal. That one angle, together with how far away you are stand­ing, is enough to work out the height of the whole struc­ture, with­out ever putting a mea­sur­ing tape on it. This is the idea behind Chap­ter 9 of the NCERT Class 10 text­book, Some Appli­ca­tions of Trigonom­e­try: we use the trigono­met­ric ratios you already know (sine, cosine, tan­gent and their rec­i­p­ro­cals) to find heights and dis­tances that are awk­ward or impos­si­ble to mea­sure directly. Sur­vey­ors, engi­neers, nav­i­ga­tors and astronomers have used exactly this method for cen­turies.

This les­son cov­ers the three basic words of the chap­ter (line of sight, angle of ele­va­tion, angle of depres­sion) and the sim­plest kind of prob­lem: one right tri­an­gle and one known angle.

The three basic ideas#

Line of sight#

Pic­ture a stu­dent stand­ing on the ground, look­ing at the top of a minaret. The straight line from the stu­den­t's eye to the top of the minaret is called the line of sight. In gen­eral, the line of sight is the line drawn from the eye of an observer to the point in the object being viewed.

Angle of ele­va­tion#

Now draw a sec­ond line from the stu­den­t's eye, straight out hor­i­zon­tally. The angle between this hor­i­zon­tal line and the line of sight is the angle of ele­va­tion of the top of the minaret. So the angle of ele­va­tion is the angle the line of sight makes with the hor­i­zon­tal when the point being viewed is above the hor­i­zon­tal level, that is, when we raise our head to look at it.

Angle of depres­sion#

Now pic­ture a girl sit­ting on a bal­cony, look­ing down at a flower pot on a step below her. Her line of sight is now below the hor­i­zon­tal. The angle between the hor­i­zon­tal through her eye and the line of sight to the pot is the angle of depres­sion of the flower pot. So the angle of depres­sion is the angle the line of sight makes with the hor­i­zon­tal when the point being viewed is below the hor­i­zon­tal level, that is, when we lower our head to look at it.

Why the two angles are equal#

These are really the same idea seen from oppo­site ends. Sup­pose a per­son at point $P$ on a bal­cony looks down at a point $Q$ on the ground, and a per­son at $Q$ looks up at $P$. The hor­i­zon­tal through $P$ and the hor­i­zon­tal ground through $Q$ are par­al­lel lines, and the line of sight $PQ$ cuts both of them. The angle of depres­sion at $P$ and the angle of ele­va­tion at $Q$ are then alter­nate angles, so they are equal:

$$\text{angle of depression of } Q \text{ from } P = \text{angle of elevation of } P \text{ from } Q$$

This fact is used again and again: when­ever a prob­lem gives an angle of depres­sion, we can trans­fer it to the bot­tom of the pic­ture and use it as an angle of ele­va­tion inside a right tri­an­gle.

Turn­ing an angle into a height#

What you need to know#

To find the height of a minaret with­out climb­ing it, you need three things:

1. the hor­i­zon­tal dis­tance from you to the foot of the minaret;
2. the angle of ele­va­tion of the top of the minaret, mea­sured at your eye;
3. the height of your eye above the ground.

Why your own height? Your eye is not at ground level. The right tri­an­gle you mea­sure has its hor­i­zon­tal side at eye level, so it only gives the part of the minaret above your eye. You add your eye height at the end to get the full height from the ground. If a prob­lem does not men­tion the observer's height, treat the observer as a point on the ground.

The key result#

In the right tri­an­gle formed by your eye, the top of the minaret and the point on the minaret at eye level, the extra height is the side oppo­site the angle of ele­va­tion $\theta$ and the hor­i­zon­tal dis­tance $d$ is the side adja­cent to it. Since $\tan \theta$ is oppo­site over adja­cent,

$$\displaystyle \tan \theta = \frac{\text{height above eye level}}{d} \quad\Rightarrow\quad \text{height above eye level} = d \tan \theta$$

$$\text{total height} = d \tan \theta + \text{eye height}$$

When the side you know or want is a slant­ing length (a lad­der, a kite string, a rope), use $\displaystyle \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$ or $\displaystyle \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ instead.

Val­ues you will use most#

$$\displaystyle \begin{aligned} \sin 30^\circ &= \tfrac{1}{2}, & \cos 30^\circ &= \tfrac{\sqrt{3}}{2}, & \tan 30^\circ &= \tfrac{1}{\sqrt{3}} \\ \sin 45^\circ &= \tfrac{1}{\sqrt{2}}, & \cos 45^\circ &= \tfrac{1}{\sqrt{2}}, & \tan 45^\circ &= 1 \\ \sin 60^\circ &= \tfrac{\sqrt{3}}{2}, & \cos 60^\circ &= \tfrac{1}{2}, & \tan 60^\circ &= \sqrt{3} \end{aligned}$$

The method, step by step#

1. Draw a neat fig­ure. Mark the hor­i­zon­tal, the ver­ti­cal object, the observer and the line of sight.
2. If an angle of depres­sion is given, trans­fer it to the lower point as an equal angle of ele­va­tion (alter­nate angles).
3. Pick out a right tri­an­gle that con­tains the known angle and the known side.
4. Label the sides as oppo­site, adja­cent and hypotenuse with respect to the known angle.
5. Choose the one ratio that links the side you know to the side you want: $\tan$ or $\cot$ for oppo­site and adja­cent, $\sin$ for oppo­site and hypotenuse, $\cos$ for adja­cent and hypotenuse.
6. Solve, then add or sub­tract any eye height, and give the answer with units (in surd form, and in dec­i­mals if the ques­tion asks).

Worked exam­ples#

Exam­ple 1: a tower and a sin­gle angle#

A tower stands ver­ti­cally on the ground. From a point on the ground 15 m away from the foot of the tower, the angle of ele­va­tion of the top of the tower is $60^\circ$. Find the height of the tower.

No eye height is given, so the observer $O$ is a point on the ground. Let $AB$ be the tower, with foot $B$. In right tri­an­gle $ABO$, $OB = 15$ m is adja­cent to the $60^\circ$ angle and $AB$ is oppo­site it, so we use tan­gent:

$$\displaystyle \tan 60^\circ = \frac{AB}{OB} \quad\Rightarrow\quad \sqrt{3} = \frac{AB}{15} \quad\Rightarrow\quad AB = 15\sqrt{3} \text{ m}$$

Since $\sqrt{3} \approx 1.732$, the height is about $25.98$ m, a lit­tle under 26 m.

Answer: the tower is $15\sqrt{3}$ m $\approx 25.98$ m tall.

Exam­ple 2: how long a lad­der does the elec­tri­cian need?#

An elec­tri­cian has to repair a fault on a pole of height 5 m. She needs to reach a point 1.3 m below the top of the pole. What should be the length of the lad­der she uses, which, when inclined at an angle of $60^\circ$ to the hor­i­zon­tal, would enable her to reach the required posi­tion? How far from the foot of the pole should she place the foot of the lad­der? (Take $\sqrt{3} = 1.73$.)

Step 1: the height to reach. $5 - 1.3 = 3.7$ m above the ground.

Step 2: the lad­der. The lad­der is the hypotenuse, and the height 3.7 m is oppo­site the $60^\circ$ angle. Oppo­site and hypotenuse call for sine:

$$\displaystyle \sin 60^\circ = \frac{3.7}{\text{ladder}} \quad\Rightarrow\quad \frac{\sqrt{3}}{2} = \frac{3.7}{\text{ladder}} \quad\Rightarrow\quad \text{ladder} = \frac{7.4}{\sqrt{3}} = \frac{7.4}{1.73} \approx 4.28 \text{ m}$$

Step 3: the foot of the lad­der. The dis­tance from the pole is adja­cent to the $60^\circ$ angle, and we know the oppo­site side, so we use cotan­gent:

$$\displaystyle \cot 60^\circ = \frac{\text{distance}}{3.7} \quad\Rightarrow\quad \frac{1}{\sqrt{3}} = \frac{\text{distance}}{3.7} \quad\Rightarrow\quad \text{distance} = \frac{3.7}{1.73} \approx 2.14 \text{ m}$$

Answer: a lad­der of about 4.28 m, with its foot about 2.14 m from the pole. (With the more pre­cise $\sqrt{3} \approx 1.732$ the val­ues are 4.27 m and 2.14 m; fol­low the value the ques­tion tells you to use.)

Exam­ple 3: a chim­ney, seen from eye level#

An observer 1.5 m tall is 28.5 m away from a chim­ney. The angle of ele­va­tion of the top of the chim­ney from her eyes is $45^\circ$. What is the height of the chim­ney?

The hor­i­zon­tal through her eye meets the chim­ney 1.5 m above the ground. In the right tri­an­gle at eye level, the hor­i­zon­tal dis­tance 28.5 m is adja­cent to $45^\circ$ and the height above eye level is oppo­site:

$$\displaystyle \tan 45^\circ = \frac{\text{height above eye}}{28.5} \quad\Rightarrow\quad 1 = \frac{\text{height above eye}}{28.5} \quad\Rightarrow\quad \text{height above eye} = 28.5 \text{ m}$$

Now add her eye height:

$$\text{height of chimney} = 28.5 + 1.5 = 30 \text{ m}$$

Answer: the chim­ney is 30 m tall.

Exam­ple 4: a boat seen from a cliff (angle of depres­sion)#

From the top of a cliff 50 m high, the angle of depres­sion of a boat on the sea is $30^\circ$. How far is the boat from the foot of the cliff?

By alter­nate angles, the angle of ele­va­tion of the top of the cliff from the boat is also $30^\circ$. In the right tri­an­gle formed by the cliff (50 m, oppo­site) and the dis­tance $d$ (adja­cent):

$$\displaystyle \tan 30^\circ = \frac{50}{d} \quad\Rightarrow\quad \frac{1}{\sqrt{3}} = \frac{50}{d} \quad\Rightarrow\quad d = 50\sqrt{3} \approx 86.60 \text{ m}$$

Answer: the boat is $50\sqrt{3}$ m $\approx 86.60$ m from the foot of the cliff.

Exam­ple 5: the height of a kite#

A kite is fly­ing on a string 60 m long, pulled tight. The string makes an angle of $60^\circ$ with the level ground. Find the height of the kite (assume there is no slack in the string).

The string is the hypotenuse and the height $h$ is oppo­site the $60^\circ$ angle, so we use sine:

$$\displaystyle \sin 60^\circ = \frac{h}{60} \quad\Rightarrow\quad h = 60 \times \frac{\sqrt{3}}{2} = 30\sqrt{3} \approx 51.96 \text{ m}$$

Answer: the kite is $30\sqrt{3}$ m $\approx 51.96$ m high.

The pat­tern to hold onto#

Every prob­lem above fol­lows the same rou­tine: draw the right tri­an­gle, decide which side is oppo­site and which is adja­cent to the known angle, choose the one ratio that links the known side to the wanted side, and solve. If the observer's eye is above the ground, add that height at the end; if an angle of depres­sion is given, move it to the lower point first. In the next les­son we meet prob­lems with two angles, such as a flag on top of a build­ing, or a shadow that changes length as the sun moves, where two right tri­an­gles must be han­dled together.

Com­mon mis­takes#

• Mea­sur­ing the angle of ele­va­tion or depres­sion from the ver­ti­cal instead of the hor­i­zon­tal.
• Plac­ing the angle of depres­sion inside the tri­an­gle at the top, between the line of sight and the ver­ti­cal. That angle is $90^\circ$ minus the angle of depres­sion, not the angle itself.
• For­get­ting to add the observer's eye height, or adding it when the ques­tion treats the observer as a point.
• Using $\sin$ when the two sides involved are the oppo­site and adja­cent sides; that pair always needs $\tan$ or $\cot$.
• Mix­ing up $\displaystyle \tan 30^\circ = \frac{1}{\sqrt{3}}$ and $\tan 60^\circ = \sqrt{3}$.
• Round­ing $\sqrt{3}$ early, or using a dif­fer­ent value from the one given in the ques­tion.

Try these#

1. From a point 30 m from the foot of a tower, the angle of ele­va­tion of its top is $30^\circ$. Find the height of the tower. Answer: $10\sqrt{3} \approx 17.32$ m.
2. A kite string 100 m long makes an angle of $60^\circ$ with the ground. Find the height of the kite. Answer: $50\sqrt{3} \approx 86.60$ m.
3. From the top of a light­house 75 m high, the angle of depres­sion of a ship is $45^\circ$. How far is the ship from the light­house? Answer: 75 m.
4. A boy whose eyes are 1.6 m above the ground stands 20 m from a tree and sees its top at an angle of ele­va­tion of $45^\circ$. Find the height of the tree. Answer: 21.6 m.
5. A 10 m lad­der leans against a wall, mak­ing an angle of $60^\circ$ with the ground. How high up the wall does it reach, and how far is its foot from the wall? Answer: $5\sqrt{3} \approx 8.66$ m up the wall; foot 5 m from the wall.

Key terms#

Line of sight

The straight line from the eye of the observer to the point on the object being viewed.

Hor­i­zon­tal

A line through the observer's eye par­al­lel to the level ground.

Angle of ele­va­tion

The angle between the line of sight and the hor­i­zon­tal when the object is above eye level.

Angle of depres­sion

The angle between the line of sight and the hor­i­zon­tal when the object is below eye level.

Alter­nate angles

Equal angles formed on oppo­site sides of a line that cuts two par­al­lel lines; they make the angle of depres­sion equal to the match­ing angle of ele­va­tion.

Oppo­site side

In a right tri­an­gle, the side fac­ing the angle being used.

Adja­cent side

In a right tri­an­gle, the side next to the angle being used, other than the hypotenuse.

Hypotenuse

The side oppo­site the right angle; the longest side, such as a lad­der or kite string.

Com­mon ques­tions#

Is the angle of ele­va­tion always equal to the angle of depres­sion?#

Between the same two points, yes. The angle of ele­va­tion of $P$ seen from $Q$ equals the angle of depres­sion of $Q$ seen from $P$, because the two hor­i­zon­tals are par­al­lel and the angles are alter­nate angles.

Which ratio should I use in a heights and dis­tances prob­lem?#

Look at the two sides involved. Height and hor­i­zon­tal dis­tance need $\tan$ (or $\cot$). A slant­ing length with the height needs $\sin$; a slant­ing length with the hor­i­zon­tal dis­tance needs $\cos$.

When do I add the observer's height?#

Only when the ques­tion gives it. The trigono­met­ric tri­an­gle then starts at eye level, so the height it gives is the part above the eye, and you add the eye height to reach the ground.

Should I leave the answer as a surd or a dec­i­mal?#

Write the exact surd form first, such as $15\sqrt{3}$ m. If the ques­tion gives a value like $\sqrt{3} = 1.73$ or $1.732$, also give the dec­i­mal using exactly that value.

Can the angle of ele­va­tion be $90^\circ$?#

Only if the object is directly over­head. Then there is no tri­an­gle, and the method can­not give a height, which is why prob­lems always use an acute angle.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Loney, S. L. Plane Trigonom­e­try. Cam­bridge Uni­ver­sity Press.
3. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
4. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.