Angle of Depression Problems with Two Buildings, Class 10

The key fact#

The hor­i­zon­tal through the observer and the hor­i­zon­tal ground are par­al­lel. The line of sight cuts both of them, so it makes a pair of alter­nate angles, and alter­nate angles between par­al­lel lines are equal. Hence:

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

For exam­ple, if the angle of depres­sion from the top of a build­ing to a point on the ground is $30^\circ$, then the angle of ele­va­tion of the top of the build­ing from that point is also $30^\circ$.

Why this mat­ters#

The angle of depres­sion sits out­side the right tri­an­gle, between the hor­i­zon­tal and the line of sight. We can­not use it directly in a ratio. The alter­nate-angle fact moves it to the lower end of the line of sight, where it is an angle inside a right tri­an­gle, next to the hor­i­zon­tal side. The angle inside the tri­an­gle at the top is $90^\circ$ minus the angle of depres­sion, which is a dif­fer­ent angle alto­gether.

The method, step by step#

1. Draw the observer's hor­i­zon­tal as a dashed line and mark each angle of depres­sion below it.
2. Trans­fer each angle to the lower end of its line of sight using alter­nate angles.
3. If a point of inter­est is above the ground (the top of a shorter build­ing), draw a hor­i­zon­tal from it to the taller struc­ture. This cre­ates a rec­tan­gle whose oppo­site sides are equal.
4. Write one $\tan$ equa­tion for each right tri­an­gle.
5. Use the rec­tan­gle or the shared ver­ti­cal to con­nect the equa­tions, then solve.
6. Check whether the required length is a sum (points on oppo­site sides) or a dif­fer­ence (points on the same side).

Worked exam­ples#

Exam­ple 1: a car seen from a tower#

From the top of a tower 60 m high, the angle of depres­sion of a car on the road is $30^\circ$. How far is the car from the foot of the tower?

By alter­nate angles, the angle of ele­va­tion of the top of the tower from the car is $30^\circ$. With the tower (60 m) oppo­site and the dis­tance $d$ adja­cent:

$$\displaystyle \tan 30^\circ = \frac{60}{d} \quad\Rightarrow\quad d = 60\sqrt{3} \approx 103.92 \text{ m}$$

Answer: the car is $60\sqrt{3}$ m $\approx 103.92$ m from the foot of the tower.

Exam­ple 2: two build­ings, two angles of depres­sion#

From the top of a multi-storeyed build­ing, the angles of depres­sion of the top and the bot­tom of an 8 m tall build­ing are $30^\circ$ and $45^\circ$ respec­tively. Find the height of the multi-storeyed build­ing and the dis­tance between the two build­ings.

Nam­ing the points. Let $PC$ be the multi-storeyed build­ing, with top $P$ and foot $C$. Let $AB$ be the 8 m build­ing, with foot $A$ and top $B$. Draw a hor­i­zon­tal from $B$ to meet $PC$ at $D$. Since $AB$ and $DC$ are ver­ti­cal and $AC$ and $BD$ are hor­i­zon­tal, $ABDC$ is a rec­tan­gle, so

$$DC = AB = 8 \text{ m} \qquad \text{and} \qquad BD = AC$$

Trans­fer­ring the angles. The angle of depres­sion of $B$ from $P$ is $30^\circ$, so $\angle PBD = 30^\circ$. The angle of depres­sion of $A$ from $P$ is $45^\circ$, so $\angle PAC = 45^\circ$.

Tri­an­gle $PBD$ (right-angled at $D$):

$$\displaystyle \tan 30^\circ = \frac{PD}{BD} \quad\Rightarrow\quad \frac{1}{\sqrt{3}} = \frac{PD}{BD} \quad\Rightarrow\quad BD = PD\sqrt{3}$$

Tri­an­gle $PAC$ (right-angled at $C$):

$$\displaystyle \tan 45^\circ = \frac{PC}{AC} \quad\Rightarrow\quad PC = AC$$

Con­nect­ing them. $PC = PD + DC = PD + 8$, and $PC = AC = BD = PD\sqrt{3}$. There­fore

$$\displaystyle \begin{aligned} PD + 8 &= PD\sqrt{3} \\ 8 &= PD(\sqrt{3} - 1) \\ PD &= \frac{8}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{8(\sqrt{3} + 1)}{3 - 1} = 4(\sqrt{3} + 1) \end{aligned}$$

So the height of the multi-storeyed build­ing is

$$PC = PD + 8 = 4\sqrt{3} + 4 + 8 = 4(3 + \sqrt{3}) \text{ m} \approx 18.93 \text{ m}$$

and since $AC = PC$, the dis­tance between the build­ings is also $4(3 + \sqrt{3})$ m.

Answer: the build­ing is $4(3 + \sqrt{3}) \approx 18.93$ m tall, and the build­ings are $4(3 + \sqrt{3}) \approx 18.93$ m apart.

Why this works: the two answers are equal because $\tan 45^\circ = 1$ makes the two legs of tri­an­gle $PAC$ equal.

Exam­ple 3: the width of a river, seen from a bridge#

From a point on a bridge across a river, the angles of depres­sion of the banks on oppo­site sides of the river are $30^\circ$ and $45^\circ$. If the bridge is at a height of 3 m from the banks, find the width of the river.

Let $A$ and $B$ be the points on the two banks, so $AB$ is the width. Let $P$ be the point on the bridge and $D$ the point on $AB$ directly below $P$, so $PD = 3$ m and $PD \perp AB$. (The height is mea­sured from the level of the banks, so $D$ lies on the line $AB$, not on the water sur­face.)

By alter­nate angles, $\angle PAD = 30^\circ$ and $\angle PBD = 45^\circ$.

Tri­an­gle $APD$:

$$\displaystyle \tan 30^\circ = \frac{PD}{AD} \quad\Rightarrow\quad \frac{1}{\sqrt{3}} = \frac{3}{AD} \quad\Rightarrow\quad AD = 3\sqrt{3} \text{ m}$$

Tri­an­gle $BPD$:

$$\displaystyle \tan 45^\circ = \frac{PD}{BD} \quad\Rightarrow\quad 1 = \frac{3}{BD} \quad\Rightarrow\quad BD = 3 \text{ m}$$

$A$ and $B$ are on oppo­site sides of $D$, so the lengths add:

$$AB = AD + DB = 3\sqrt{3} + 3 = 3(\sqrt{3} + 1) \text{ m} \approx 8.20 \text{ m}$$

Answer: the river is $3(\sqrt{3} + 1)$ m $\approx 8.20$ m wide.

Exam­ple 4: two cars on the same side#

From the top of a tower 100 m high, a man observes two cars on the same side of the tower, in a straight line with its foot. The angles of depres­sion of the cars are $30^\circ$ and $45^\circ$. Find the dis­tance between the cars.

By alter­nate angles the angles of ele­va­tion from the cars are $30^\circ$ and $45^\circ$. The nearer car (larger angle) is at dis­tance

$$d_1 = 100 \cot 45^\circ = 100 \text{ m}$$

and the far­ther car is at dis­tance

$$d_2 = 100 \cot 30^\circ = 100\sqrt{3} \text{ m}$$

Both cars are on the same side, so the dis­tances sub­tract:

$$d_2 - d_1 = 100\sqrt{3} - 100 = 100(\sqrt{3} - 1) \approx 73.2 \text{ m}$$

Answer: the cars are $100(\sqrt{3} - 1)$ m $\approx 73.2$ m apart.

Exam­ple 5: ele­va­tion and depres­sion from the same roof#

From the top of a 7 m high build­ing, the angle of ele­va­tion of the top of a cable tower is $60^\circ$ and the angle of depres­sion of its foot is $45^\circ$. Find the height of the tower.

Let the build­ing be $AB$ with top $A$, and the tower $CE$ with foot $C$ and top $E$. Draw the hor­i­zon­tal $AF$ from $A$ to meet the tower at $F$, so $FC = AB = 7$ m and $AF = BC$.

From the angle of depres­sion, $\angle ACB = 45^\circ$ in tri­an­gle $ABC$:

$$\displaystyle \tan 45^\circ = \frac{AB}{BC} \quad\Rightarrow\quad BC = 7 \text{ m, so } AF = 7 \text{ m}$$

In tri­an­gle $AFE$, with $\angle EAF = 60^\circ$:

$$\displaystyle \tan 60^\circ = \frac{EF}{AF} \quad\Rightarrow\quad EF = 7\sqrt{3} \text{ m}$$

$$CE = EF + FC = 7\sqrt{3} + 7 = 7(\sqrt{3} + 1) \text{ m} \approx 19.12 \text{ m}$$

Answer: the tower is $7(\sqrt{3} + 1) \approx 19.12$ m high.

Bring­ing it all together#

The same few moves appear in every prob­lem: draw and label the right tri­an­gles care­fully; con­vert each angle of depres­sion into an equal angle of ele­va­tion using alter­nate angles; draw a hor­i­zon­tal through any raised point to cre­ate a rec­tan­gle; iden­tify what the tri­an­gles share; and write one equa­tion per tri­an­gle before com­bin­ing them. The trigono­met­ric ratios them­selves are the ones you already know. What this chap­ter adds is the prac­ti­cal eye for turn­ing a tower, a lad­der, a flagstaff or a river into the right tri­an­gle that solves it.

Com­mon mis­takes#

• Mark­ing the angle of depres­sion between the line of sight and the ver­ti­cal wall of the build­ing.
• Using the full height $PC$ in the tri­an­gle for the top of the shorter build­ing; that tri­an­gle only con­tains $PD = PC - 8$.
• Adding the dis­tances when both points are on the same side of the tower (they should be sub­tracted), or sub­tract­ing when they are on oppo­site sides.
• Assum­ing the larger angle of depres­sion belongs to the far­ther point; a larger angle means the point is nearer.
• For­get­ting that the rec­tan­gle gives $BD = AC$, which is the link between the two tri­an­gles.
• Stop­ping at $\displaystyle \frac{8}{\sqrt{3} - 1}$ with­out ratio­nal­is­ing and sim­pli­fy­ing.

Try these#

1. From the top of a light­house 100 m high, the angle of depres­sion of a boat is $30^\circ$. How far is the boat from the light­house? Answer: $100\sqrt{3} \approx 173.21$ m.
2. From the top of a 75 m high light­house, the angles of depres­sion of two ships on oppo­site sides of it are $30^\circ$ and $45^\circ$. Find the dis­tance between the ships. Answer: $75(\sqrt{3} + 1) \approx 204.90$ m.
3. From the top of a 60 m high build­ing, the angles of depres­sion of the top and the bot­tom of a tower are $30^\circ$ and $60^\circ$. Find the height of the tower. Answer: 40 m.
4. A bridge is 5 m above the banks of a river. From a point on it, the angles of depres­sion of the two oppo­site banks are $60^\circ$ and $30^\circ$. Find the width of the river. Answer: $\displaystyle \frac{20\sqrt{3}}{3} \approx 11.55$ m.
5. From the top of a 20 m high build­ing, the angle of ele­va­tion of the top of a tower is $60^\circ$ and the angle of depres­sion of its foot is $45^\circ$. Find the height of the tower. Answer: $20(\sqrt{3} + 1) \approx 54.64$ m.

Key terms#

Angle of depres­sion

The angle between the hor­i­zon­tal through the observer's eye and the line of sight to a point below it.

Alter­nate angles

Angles on oppo­site sides of a trans­ver­sal between two par­al­lel lines; they are equal.

Trans­ver­sal

A line that cuts two or more other lines; here, the line of sight cross­ing two hor­i­zon­tals.

Hor­i­zon­tal through a point

A line drawn through the point par­al­lel to the ground, used to mea­sure angles of ele­va­tion and depres­sion.

Rec­tan­gle con­struc­tion

Draw­ing a hor­i­zon­tal from the top of the shorter struc­ture to the taller one so that oppo­site sides become equal.

Width of a river

The hor­i­zon­tal dis­tance between points on oppo­site banks, found as a sum of two hor­i­zon­tal pieces.

Ratio­nal­is­ing fac­tor

An expres­sion such as $\sqrt{3} + 1$ that removes a surd from the denom­i­na­tor $\sqrt{3} - 1$.

Com­mon ques­tions#

Why can't I use the angle of depres­sion directly in the tri­an­gle?#

It lies between the hor­i­zon­tal and the line of sight, out­side the right tri­an­gle. Alter­nate angles move it to the lower point, where it lies inside the tri­an­gle next to the hor­i­zon­tal side.

How do I tell whether to add or sub­tract the dis­tances?#

Draw the fig­ure. If the two points lie on oppo­site sides of the foot of the per­pen­dic­u­lar, add; if they are on the same side, sub­tract the shorter dis­tance from the longer one.

Which point is nearer, the one with the larger or smaller angle of depres­sion?#

The larger angle of depres­sion belongs to the nearer point, because the dis­tance is height $\times \cot \theta$ and $\cot \theta$ decreases as $\theta$ increases.

Why are the height and dis­tance equal in the two-build­ings prob­lem?#

The foot of the shorter build­ing is seen at $45^\circ$, and $\tan 45^\circ = 1$, so the ver­ti­cal and hor­i­zon­tal legs of that tri­an­gle are equal.

Do I always need the rec­tan­gle?#

Only when one of the points is above the ground, such as the top of a shorter build­ing. The rec­tan­gle turns the raised point into a right tri­an­gle with a known ver­ti­cal side.

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. Heath, T. L. (trans.) The Thir­teen Books of Euclid's Ele­ments. Dover Pub­li­ca­tions.