You Can Measure a Tower You Cannot Climb With One Angle and One Distance

Stand $30$ m from the foot of a tower, look up at its top, and measure the angle your line of sight makes with the horizontal. Suppose it is $30^{\circ }$. That is enough to give you the height of the tower to the nearest centimetre, and you never have to leave the ground.

The reason is the ratio you learnt in the previous two chapters. The tower, the ground and your line of sight form a right triangle, and the tangent of the angle is the height divided by the horizontal distance:

$$\tan 30^{\circ }=\frac{h}{30}⟹h=30\times \frac{1}{\sqrt{3}}=10\sqrt{3}\approx 17.32\text{ m}$$

One angle, one distance, one line of working. That is the whole of this chapter, applied over and over to taller and more awkward situations.

What changes from question to question is only the diagram. And that is where the marks live:

- Which angle is given, and whether it is measured up from the horizontal or down from it
- Where the right angle is, which decides which side is the hypotenuse
- How many triangles the situation contains, because the hardest questions hide two

So the chapter is much less about trigonometry than it looks. You already know every ratio you need. The skill being built is reading a written situation and turning it into a correctly labelled figure — and a student who draws the figure carefully almost never gets the arithmetic wrong afterwards.

The angles used are restricted to $30^{\circ }$, $45^{\circ }$ and $60^{\circ }$, whose exact values you know, so no table or calculator is needed. **Where a decimal answer is wanted, take $\sqrt{3}=1.732$.**

This page covers the CBSE Class 10 Maths chapter on some applications of trigonometry: the line of sight, the angles of elevation and depression, and height-and-distance problems with one and two triangles.

The line of sight is the straight line from the observer's eye to the object. The angle of elevation is the angle it makes with the horizontal when the object is above the eye; the angle of depression is that angle when the object is below.

Both are always measured from the horizontal, never from the vertical. That single sentence prevents most of the errors in this chapter.

How to draw the figure for an elevation. The observer stands at a point on the ground; the object is above. Draw:

- A horizontal line from the observer's eye
- A vertical line for the object's height, meeting the ground at a right angle
- The line of sight from the eye to the top of the object
- The angle between the horizontal and the line of sight, marked at the observer

The right angle is always at the foot of the vertical object, and the line of sight is always the hypotenuse. Marking the right angle on your figure is worth a mark on its own.

How to draw the figure for a depression. The observer is now at the top of a tower, a cliff or an aircraft, and the object is below. Draw a horizontal line at the observer's level — this is the essential step — and measure the angle down from it to the line of sight.

And now the single most useful fact in the chapter. The horizontal at the observer's level and the ground are parallel, and the line of sight crosses both. So the angle of depression from the top equals the angle of elevation from the bottom, as alternate angles between parallel lines.

That means you never have to work with a depression as such. Redraw it as an elevation from the object looking up, and the triangle becomes the same shape you are used to.

Worked example — reading a situation. A boy $1.5$ m tall stands $20$ m from the foot of a lamp post and looks up at the lamp, which is $9$ m above the ground. Find the angle of elevation.

The height that matters is the height above the boy's eye, not above the ground:

$$9-1.5=7.5\text{ m}$$

But wait — the tangent would then be $\frac{7.5}{20}=0.375$, which is not one of the standard values. So this situation does not have a standard angle, and a question set at this level would choose the numbers so that it does. **That is itself a useful check: if your tangent is not $\frac{1}{\sqrt{3}}$, $1$ or $\sqrt{3}$, re-read the question.

Worked example 2 — where the eye level is given and does matter.** A boy $1.5$ m tall observes the top of a building at an elevation of $45^{\circ }$ from a point $28.5$ m from the building. Find the height of the building.

The height above his eye is

$$28.5\times \tan 45^{\circ }=28.5\text{ m}$$

and the total height is

$$28.5+1.5=30\text{ m}$$

The observer's own height had to be added back at the end. Forgetting it is the commonest slip when a question bothers to mention how tall the observer is — and if the question does not mention it, the observer is taken to be at ground level.

One boundary case worth noticing. At $45^{\circ }$ the tangent is $1$, so the height equals the horizontal distance. Any question involving $45^{\circ }$ therefore hands you a length for free, and in a two-triangle problem that is almost always the triangle to start with.

Decide which two sides the question involves, pick the ratio that connects them, substitute, and solve. Height and horizontal distance means tangent; height and line of sight means sine; horizontal distance and line of sight means cosine.

Worked example 1 — the height, from the distance. The angle of elevation of the top of a tower from a point on the ground $30$ m from its foot is $30^{\circ }$. Find the height of the tower.

The two sides involved are the opposite (the height) and the adjacent (the ground distance), so use the tangent:

$$\tan 30^{\circ }=\frac{h}{30}⟹\frac{1}{\sqrt{3}}=\frac{h}{30}$$
$$h=\frac{30}{\sqrt{3}}=\frac{30\sqrt{3}}{3}=10\sqrt{3}\approx 17.32\text{ m}$$

Check the size of the answer. At $30^{\circ }$ the line of sight is fairly flat, so the tower should be much shorter than $30$ m — and $17.32$ m is. A sanity check on the magnitude catches a flipped ratio instantly, because writing $\tan 30^{\circ }=\frac{30}{h}$ would have given $h=51.96$ m, taller than the distance, which is impossible at $30^{\circ }$.

Worked example 2 — the distance, from the height. The angle of elevation of the top of a $75$ m tower from a point on the ground is $60^{\circ }$. How far is the point from the foot of the tower?

$$\tan 60^{\circ }=\frac{75}{d}⟹\sqrt{3}=\frac{75}{d}$$
$$d=\frac{75}{\sqrt{3}}=25\sqrt{3}\approx 43.3\text{ m}$$

And the magnitude is right: at $60^{\circ }$ the sight line is steep, so the point must be much closer than the tower is tall.

Worked example 3 — where the hypotenuse is given. A ladder leans against a wall making an angle of $60^{\circ }$ with the ground, with its foot $2.5$ m from the wall. Find the length of the ladder.

The ladder is the hypotenuse and $2.5$ m is the adjacent side, so use the cosine:

$$\cos 60^{\circ }=\frac{2.5}{L}⟹\frac{1}{2}=\frac{2.5}{L}⟹L=5\text{ m}$$

Check: the ladder must be longer than the distance from the wall, and $5>2.5$. Correct.

Worked example 4 — where the hypotenuse is given and the height is wanted. The string of a kite is $100$ m long and makes an angle of $60^{\circ }$ with the horizontal ground. Find the height of the kite, assuming the string is straight.

Now the string is the hypotenuse and the height is the opposite side, so use the sine:

$$\sin 60^{\circ }=\frac{h}{100}⟹h=100\times \frac{\sqrt{3}}{2}=50\sqrt{3}\approx 86.6\text{ m}$$

Check: the height cannot exceed the length of the string, and $86.6<100$. Correct, and if your answer for a height ever comes out larger than the hypotenuse, a ratio has been inverted.

Worked example 5 — a broken tree, which needs two ratios. A tree breaks in a storm and the broken part bends so that its top touches the ground, making an angle of $30^{\circ }$ with the ground. The distance from the foot of the tree to the point where the top touches the ground is $8$ m. Find the original height of the tree.

The broken part is now the hypotenuse and the standing part is the vertical side. Call the standing part $a$ and the broken part $b$.

For the standing part, use the tangent:

$$\tan 30^{\circ }=\frac{a}{8}⟹a=\frac{8}{\sqrt{3}}$$

For the broken part, use the cosine:

$$\cos 30^{\circ }=\frac{8}{b}⟹b=\frac{8}{\cos 30^{\circ }}=\frac{16}{\sqrt{3}}$$

The original height is the standing part plus the broken part:

$$a+b=\frac{8}{\sqrt{3}}+\frac{16}{\sqrt{3}}=\frac{24}{\sqrt{3}}=8\sqrt{3}\approx 13.86\text{ m}$$

Check with the Pythagoras theorem: $a=\frac{8}{\sqrt{3}}\approx 4.62$ and $b=\frac{16}{\sqrt{3}}\approx 9.24$, and $4.62^2+8^2\approx 21.33+64=85.33$, while $9.24^2\approx 85.33$. The two agree, so the two ratios were applied consistently.

The idea worth extracting. The tree's original height is not the hypotenuse and not the standing part — it is their sum, because the broken piece used to stand on top of the stump. Reading what quantity the question actually wants is half the difficulty in this chapter, and it is why the figure must be drawn before any ratio is chosen.

Draw the horizontal at the observer's level, mark the depression below it, and then use the fact that it equals the angle of elevation from the object. After that the working is identical to an elevation problem.

Worked example 1 — a car from a tower. From the top of a $75$ m tall tower, the angle of depression of a car standing on the ground is $30^{\circ }$. Find the distance of the car from the foot of the tower.

The angle of depression from the top is $30^{\circ }$, so **the angle of elevation of the top from the car is also $30^{\circ }$**, being alternate angles. Then

$$\tan 30^{\circ }=\frac{75}{d}⟹d=75\sqrt{3}\approx 129.9\text{ m}$$

Check the magnitude. A shallow depression of $30^{\circ }$ means the car is far away, and $129.9$ m is comfortably more than the $75$ m height. Correct.

The error this example is built to catch. Writing $\tan 30^{\circ }=\frac{d}{75}$ gives $d=43.3$ m, which would place the car closer than the tower is tall — impossible at $30^{\circ }$. The height is opposite the angle of elevation at the car and the distance is adjacent to it, and drawing the triangle with the angle at the car rather than at the top makes this automatic.

Worked example 2 — two ships from a lighthouse. From the top of a $100$ m high lighthouse, the angles of depression of two ships on the same side are $30^{\circ }$ and $45^{\circ }$. Find the distance between the ships.

Two triangles share the same vertical side. Take the nearer ship first, at $45^{\circ }$:

$$\tan 45^{\circ }=\frac{100}{d_1}⟹d_1=100\text{ m}$$

Then the farther ship, at $30^{\circ }$:

$$\tan 30^{\circ }=\frac{100}{d_2}⟹d_2=100\sqrt{3}\approx 173.2\text{ m}$$

The distance between them is the difference:

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

Which ship is nearer, and how to be sure. A larger angle of depression means a steeper line of sight and therefore a nearer object. So the ship at $45^{\circ }$ is the nearer one, and subtracting the smaller distance from the larger is the right way round. Getting this backwards produces a negative distance, which is the check that catches it.

Worked example 3 — an aircraft. The angle of depression of a point on the ground from an aeroplane flying at a height of $1500$ m is $60^{\circ }$. Find the horizontal distance of the point from the aeroplane's position on the ground directly below it.

$$\tan 60^{\circ }=\frac{1500}{d}⟹d=\frac{1500}{\sqrt{3}}=500\sqrt{3}\approx 866\text{ m}$$

Check: a steep $60^{\circ }$ depression should give a horizontal distance smaller than the height, and $866<1500$. Correct.

Worked example 4 — a cliff and a boat approaching. From the top of a cliff $60$ m high, the angle of depression of a boat is $45^{\circ }$. A little later it is $60^{\circ }$. How much closer has the boat come?

At $45^{\circ }$: $d_1=\frac{60}{\tan 45^{\circ }}=60$ m.

At $60^{\circ }$: $d_2=\frac{60}{\sqrt{3}}=20\sqrt{3}\approx 34.64$ m.

So the boat has moved

$$60-20\sqrt{3}\approx 25.36\text{ m}$$

closer to the cliff.

Notice how the two triangles were kept separate. Each was solved on its own, using the same vertical height, and only the final step combined them. Never try to work with the small triangle between the two lines of sight — it is not right-angled, and none of the ratios of this chapter apply to it.

**Find the triangle that gives you a length immediately — almost always the one with $45^{\circ }$ — use it to get the shared side, then carry that length into the second triangle.

The hardest questions in this chapter all have the same shape: two angles and one unknown shared side. Once you see that the two triangles have a side in common, the method writes itself.

Worked example 1 — a tower on a building.** From a point on the ground, the angle of elevation of the bottom of a transmission tower fixed on the top of a $20$ m high building is $45^{\circ }$, and that of the top of the tower is $60^{\circ }$. Find the height of the tower.

Step 1 — the easy triangle. The bottom of the tower is the top of the building, at $20$ m, seen at $45^{\circ }$:

$$\tan 45^{\circ }=\frac{20}{d}⟹d=20\text{ m}$$

Step 2 — the same distance in the bigger triangle. The top of the tower is at height $20+t$, seen at $60^{\circ }$ from the same point:

$$\tan 60^{\circ }=\frac{20+t}{20}⟹20+t=20\sqrt{3}$$
$$t=20\sqrt{3}-20=20(\sqrt{3}-1)\approx 14.64\text{ m}$$

**The tower is about $14.64$ m tall.

Check**: the total height is $20+14.64=34.64$ m, and $\tan 60^{\circ }$ should be $\frac{34.64}{20}=1.732$. It is.

**The shared side was the horizontal distance $d$**, and it was found from the $45^{\circ }$ triangle because there the tangent is $1$. **That is why $45^{\circ }$ is the one to start with whenever it appears.

Worked example 2 — a flagstaff on a tower, where the shared side is unknown.** A vertical tower is surmounted by a flagstaff of height $5$ m. From a point on the ground, the angle of elevation of the bottom of the flagstaff is $30^{\circ }$ and that of its top is $45^{\circ }$. Find the height of the tower.

Here neither triangle gives a length straight away, so call the tower's height $h$ and the distance $d$ and write both equations.

From the $30^{\circ }$ triangle:

$$\tan 30^{\circ }=\frac{h}{d}⟹d=h\sqrt{3}$$

From the $45^{\circ }$ triangle:

$$\tan 45^{\circ }=\frac{h+5}{d}⟹d=h+5$$

**The shared side $d$ now links them:**

$$h\sqrt{3}=h+5⟹h(\sqrt{3}-1)=5⟹h=\frac{5}{\sqrt{3}-1}$$

Rationalising by multiplying top and bottom by $\sqrt{3}+1$:

$$h=\frac{5(\sqrt{3}+1)}{3-1}=\frac{5(\sqrt{3}+1)}{2}\approx \frac{5\times 2.732}{2}\approx 6.83\text{ m}$$

Check: $d=h+5\approx 11.83$ m, and $\tan 30^{\circ }$ should be $\frac{6.83}{11.83}=0.577$. It is.

That is the general method, and it is worth stating as a recipe: write one equation per triangle in the same two unknowns, then eliminate the shared side. It is the simultaneous-equation technique of Chapter 3 doing geometry.

Worked example 3 — a building and a cable tower. 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.

Step 1 — the depression gives the horizontal distance. The foot of the tower is $7$ m below the observer's level, at a depression of $45^{\circ }$:

$$\tan 45^{\circ }=\frac{7}{d}⟹d=7\text{ m}$$

Step 2 — the elevation gives the part above the observer's level:

$$\tan 60^{\circ }=\frac{x}{7}⟹x=7\sqrt{3}\approx 12.12\text{ m}$$

Step 3 — add the two parts:

$$7+7\sqrt{3}=7(1+\sqrt{3})\approx 7\times 2.732\approx 19.12\text{ m}$$

**The tower is about $19.12$ m tall.

The step students miss is the third one. The elevation triangle only measures the part of the tower above the roof**; the $7$ m below the roof must be added back. Drawing the horizontal through the observer's eye and marking both parts of the tower on the figure is what makes this obvious.

Worked example 4 — a shadow lengthening. 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.

Two triangles again, both with the tower as the vertical side. Let the height be $h$.

At $60^{\circ }$ the shadow is $\frac{h}{\tan 60^{\circ }}=\frac{h}{\sqrt{3}}$; at $30^{\circ }$ it is $\frac{h}{\tan 30^{\circ }}=h\sqrt{3}$. The difference is $40$ m:

$$h\sqrt{3}-\frac{h}{\sqrt{3}}=40$$

Multiplying through by $\sqrt{3}$:

$$3h-h=40\sqrt{3}⟹2h=40\sqrt{3}⟹h=20\sqrt{3}\approx 34.64\text{ m}$$

Check with actual shadow lengths. At $60^{\circ }$: $\frac{20\sqrt{3}}{\sqrt{3}}=20$ m. At $30^{\circ }$: $20\sqrt{3}\times \sqrt{3}=60$ m. **The difference is exactly $40$ m. That check is worth doing because it confirms which altitude gives the longer shadow — a lower sun gives a longer shadow**, and setting up the subtraction the other way round would have produced a negative height.