Heights and Distances: Problems with Two Angles of Elevation

In the pre­vi­ous les­son every prob­lem gave one right tri­an­gle and one known angle, and that was enough to find the one unknown. Real sit­u­a­tions are often richer. Stand­ing at one spot, you might mea­sure the angle of ele­va­tion of the top of a build­ing and the angle of ele­va­tion of the top of a flag fly­ing above it. Or you might mea­sure the same tow­er's shadow at two dif­fer­ent times of day. Either way you get two right tri­an­gles that share a side, and two unknowns instead of one. This les­son shows how to link the two tri­an­gles, using the NCERT Class 10 exam­ples from Chap­ter 9 and a few more of the same type that appear often in board exam­i­na­tions.

The idea: two tri­an­gles, one shared quan­tity#

What makes these prob­lems dif­fer­ent#

A two-angle prob­lem has two right tri­an­gles drawn on the same fig­ure. Each tri­an­gle on its own has too lit­tle infor­ma­tion, but the two tri­an­gles always have some­thing in com­mon: a side they both use, such as the hor­i­zon­tal dis­tance from a fixed observer, or a length that does not change, such as the height of a tower.

The key result#

If a ver­ti­cal object of height $h$ is seen from a point at hor­i­zon­tal dis­tance $d$ with angle of ele­va­tion $\theta$, then

$$h = d \tan \theta \qquad \text{and} \qquad d = h \cot \theta$$

Writ­ing one such equa­tion for each tri­an­gle gives two equa­tions. Because they share a quan­tity, the equa­tions can be com­bined (by sub­sti­tu­tion or by equat­ing) to find both unknowns.

Why this works#

Each tri­an­gle is right-angled at the foot of the ver­ti­cal object, so $\tan$ of the known angle is fixed by the shape of that tri­an­gle. The shared side is lit­er­ally the same seg­ment in both tri­an­gles, so its length must sat­isfy both equa­tions at once. Two inde­pen­dent equa­tions in two unknowns can be solved, just as in a pair of lin­ear equa­tions.

The method, step by step#

1. Draw one fig­ure con­tain­ing both lines of sight. Name every point.
2. Iden­tify the two right tri­an­gles and the side or length they share.
3. Let the unknown lengths be let­ters such as $x$ and $h$.
4. Write one trigono­met­ric equa­tion for each tri­an­gle, using $\tan$ (height and hor­i­zon­tal dis­tance).
5. If one tri­an­gle has a known side, solve it first and carry the result into the sec­ond. Oth­er­wise equate the two expres­sions for the shared quan­tity.
6. Ratio­nalise any denom­i­na­tor, give the exact answer, then a dec­i­mal if asked.

Worked exam­ples#

Exam­ple 1: the build­ing and the flagstaff#

From a point $P$ on the ground, the angle of ele­va­tion of the top of a 10 m tall build­ing is $30^\circ$. A flag is hoisted at the top of the build­ing, and the angle of ele­va­tion of the top of the flagstaff from $P$ is $45^\circ$. Find the length of the flagstaff and the dis­tance of the build­ing from $P$. (Take $\sqrt{3} = 1.732$.)

Let $A$ be the foot of the build­ing, $B$ its top and $D$ the top of the flagstaff, so $AB = 10$ m and $BD$ is the flagstaff. The two right tri­an­gles $PAB$ (angle $30^\circ$) and $PAD$ (angle $45^\circ$) share the side $AP$.

Step 1: solve the tri­an­gle with a known side. In tri­an­gle $PAB$, $AB = 10$ m is oppo­site $30^\circ$ and $AP$ is adja­cent:

$$\displaystyle \tan 30^\circ = \frac{AB}{AP} \quad\Rightarrow\quad \frac{1}{\sqrt{3}} = \frac{10}{AP} \quad\Rightarrow\quad AP = 10\sqrt{3} \text{ m} \approx 17.32 \text{ m}$$

Step 2: use the sec­ond tri­an­gle. Let $BD = x$ m, so $AD = 10 + x$. In tri­an­gle $PAD$:

$$\displaystyle \tan 45^\circ = \frac{AD}{AP} \quad\Rightarrow\quad 1 = \frac{10 + x}{10\sqrt{3}} \quad\Rightarrow\quad 10 + x = 10\sqrt{3}$$

$$x = 10\sqrt{3} - 10 = 10(\sqrt{3} - 1) = 10 \times 0.732 = 7.32 \text{ m}$$

Answer: the flagstaff is $10(\sqrt{3} - 1) \approx 7.32$ m long, and the build­ing is $10\sqrt{3} \approx 17.32$ m from $P$.

Why this works: $P$ never moves, so the hor­i­zon­tal dis­tance $AP$ is the same in both tri­an­gles; only the angle changes as we look higher.

Exam­ple 2: a shadow that grows as the sun sinks#

The shadow of a tower stand­ing on level ground is found to be 40 m longer when the sun's alti­tude is $30^\circ$ than when it is $60^\circ$. Find the height of the tower.

This time the observer is not fixed. The two tri­an­gles come from the same tower at two dif­fer­ent times of day. The sun's alti­tude is the angle of ele­va­tion of the sun, which is the angle at the tip of the shadow. When the sun is higher ($60^\circ$) the shadow is shorter; when it sinks to $30^\circ$ the shadow is longer. What stays the same is the height of the tower.

Let $AB = h$ m be the tower, $BC = x$ m the shadow at $60^\circ$, and $BD = (x + 40)$ m the shadow at $30^\circ$.

Tri­an­gle $ABC$ (sun at $60^\circ$):

$$\displaystyle \tan 60^\circ = \frac{h}{x} \quad\Rightarrow\quad h = \sqrt{3}\,x \qquad \text{(1)}$$

Tri­an­gle $ABD$ (sun at $30^\circ$):

$$\displaystyle \tan 30^\circ = \frac{h}{x + 40} \quad\Rightarrow\quad h = \frac{x + 40}{\sqrt{3}} \qquad \text{(2)}$$

Both equa­tions describe the same $h$, so equate them and mul­ti­ply by $\sqrt{3}$:

$$\displaystyle \begin{aligned} \sqrt{3}\,x &= \frac{x + 40}{\sqrt{3}} \\ 3x &= x + 40 \\ 2x &= 40 \\ x &= 20 \end{aligned}$$

Sub­sti­tute in (1):

$$h = 20\sqrt{3} \text{ m} \approx 34.64 \text{ m}$$

Check with (2): $\displaystyle \frac{20 + 40}{\sqrt{3}} = \frac{60}{\sqrt{3}} = 20\sqrt{3}$, which agrees.

Answer: the tower is $20\sqrt{3}$ m $\approx 34.64$ m tall.

Exam­ple 3: walk­ing towards a tower#

From a point on level ground, the angle of ele­va­tion of the top of a tower is $30^\circ$. After walk­ing 20 m towards the tower, the angle of ele­va­tion becomes $60^\circ$. Find the height of the tower and the dis­tance of the sec­ond point from the tower.

This is the shadow prob­lem in a dif­fer­ent cos­tume. Let the height be $h$ and the dis­tance of the nearer point be $x$, so the far­ther point is $(x + 20)$ m away.

$$\displaystyle \tan 60^\circ = \frac{h}{x} \Rightarrow h = \sqrt{3}\,x, \qquad \tan 30^\circ = \frac{h}{x + 20} \Rightarrow h = \frac{x + 20}{\sqrt{3}}$$

Equat­ing: $3x = x + 20$, so $x = 10$ and $h = 10\sqrt{3}$.

Answer: the tower is $10\sqrt{3} \approx 17.32$ m tall, and the sec­ond point is 10 m from its foot.

Why this works: for $30^\circ$ and $60^\circ$ the far dis­tance is always three times the near one (since $\cot 30^\circ = 3 \cot 60^\circ$), so the 20 m walked is twice the near dis­tance.

Exam­ple 4: a statue on a pedestal#

A statue 1.6 m tall stands on top of a pedestal. From a point on the ground, the angle of ele­va­tion of the top of the statue is $60^\circ$ and from the same point the angle of ele­va­tion of the top of the pedestal is $45^\circ$. Find the height of the pedestal.

Let the pedestal have height $h$ and let the point be $d$ m from its foot. The two tri­an­gles share $d$.

$$\displaystyle \tan 45^\circ = \frac{h}{d} \quad\Rightarrow\quad d = h$$

$$\displaystyle \tan 60^\circ = \frac{h + 1.6}{d} \quad\Rightarrow\quad \sqrt{3}\,h = h + 1.6$$

$$\displaystyle \begin{aligned} h(\sqrt{3} - 1) &= 1.6 \\ h &= \frac{1.6}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{1.6(\sqrt{3} + 1)}{2} = 0.8(\sqrt{3} + 1) \end{aligned}$$

Answer: the pedestal is $0.8(\sqrt{3} + 1)$ m $\approx 2.19$ m high.

The idea to take away#

Each of these prob­lems seemed to have too lit­tle infor­ma­tion: two unknowns and only one tri­an­gle's worth of num­bers at a time. The way through was always the same. Find a quan­tity that stays fixed across both tri­an­gles (the dis­tance $AP$ for the flagstaff, the height $h$ for the shadow and the walk­ing observer, the dis­tance $d$ for the pedestal), write an equa­tion for it from each tri­an­gle, and com­bine the two. In the next les­son the same idea appears with angles of depres­sion, where the observer is high up and looks down at two points or two struc­tures.

Com­mon mis­takes#

• Using the total height $AD$ in the $30^\circ$ tri­an­gle, or the build­ing height $AB$ in the $45^\circ$ tri­an­gle; each angle belongs to its own line of sight.
• Tak­ing the sec­ond shadow as 40 m instead of $(x + 40)$ m; the 40 m is the extra length.
• Putting the sun's alti­tude at the top of the tower instead of at the tip of the shadow.
• For­get­ting that the shared side really is the same length in both tri­an­gles and intro­duc­ing two dif­fer­ent let­ters for it.
• Leav­ing $\displaystyle \frac{1.6}{\sqrt{3} - 1}$ unra­tionalised, which makes the dec­i­mal harder to com­pute cor­rectly.
• Sub­sti­tut­ing a dec­i­mal for $\sqrt{3}$ before solv­ing, which intro­duces round­ing errors into both answers.

Try these#

1. From a point on the ground, the angles of ele­va­tion of the bot­tom and top of a trans­mis­sion tower fixed on top of a 20 m high build­ing are $45^\circ$ and $60^\circ$. Find the height of the tower. Answer: $20(\sqrt{3} - 1) \approx 14.64$ m.
2. The shadow of a pole is 20 m longer when the sun's alti­tude is $30^\circ$ than when it is $60^\circ$. Find the height of the pole. Answer: $10\sqrt{3} \approx 17.32$ m.
3. The angle of ele­va­tion of the top of a tower changes from $30^\circ$ to $60^\circ$ as an observer walks 50 m towards it. Find the height of the tower. Answer: $25\sqrt{3} \approx 43.30$ m.
4. The angle of ele­va­tion of the top of a tower changes from $45^\circ$ to $60^\circ$ as an observer walks 10 m towards it. Find the height of the tower. Answer: $5(3 + \sqrt{3}) \approx 23.66$ m.
5. A tow­er's shadow is $\sqrt{3}$ times its height. Find the sun's alti­tude. Answer: $30^\circ$.

Key terms#

Angle of ele­va­tion

The angle between the hor­i­zon­tal and the line of sight to a point above the observer's eye level.

Sun's alti­tude

The angle of ele­va­tion of the sun above the hori­zon; it is the angle at the tip of a shadow.

Shadow

The hor­i­zon­tal side of the right tri­an­gle formed by a ver­ti­cal object and the sun's ray through its top.

Shared side

A length that appears in both right tri­an­gles of a fig­ure and links their equa­tions.

Flagstaff

The pole on which a flag is flown; in prob­lems it stands ver­ti­cally on top of a build­ing.

Ratio­nal­is­ing the denom­i­na­tor

Mul­ti­ply­ing numer­a­tor and denom­i­na­tor by a con­ju­gate such as $\sqrt{3} + 1$ to remove a surd from the denom­i­na­tor.

Pair of equa­tions

Two equa­tions in the same two unknowns, one from each tri­an­gle, solved together.

Com­mon ques­tions#

How do I know which tri­an­gle to solve first?#

Start with the tri­an­gle that has a known side as well as a known angle. In the flagstaff prob­lem that is tri­an­gle $PAB$, where $AB = 10$ m. If nei­ther tri­an­gle has a known side, write both equa­tions and equate the shared quan­tity.

Why does the shadow get longer when the sun is lower?#

For a fixed height $h$, the shadow is $h \cot \theta$. As the alti­tude $\theta$ decreases from $60^\circ$ to $30^\circ$, $\cot \theta$ increases from $\displaystyle \frac{1}{\sqrt{3}}$ to $\sqrt{3}$, so the shadow becomes three times as long.

Can I use $\cot$ instead of $\tan$?#

Yes. $\displaystyle \cot \theta = \frac{\text{adjacent}}{\text{opposite}}$ is often neater when the unknown is a hor­i­zon­tal dis­tance, as in $d = h \cot \theta$. Both give the same answer.

Is the walk­ing-observer prob­lem really the same as the shadow prob­lem?#

Yes. In both, one height is seen at two angles from two points on the same line, and the gap between the points is given. The alge­bra is iden­ti­cal.

How many dec­i­mal places should I give?#

Give the exact form first. Then use the value of $\sqrt{3}$ given in the ques­tion and round to two dec­i­mal places unless told oth­er­wise.

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. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.