Heights and distances

A surveyor standing 20 metres from a tower, or a lighthouse keeper watching a ship approach, both face the same puzzle: how do you measure a height or a distance you cannot reach with a tape? Heights and distances is the chapter where the trigonometric ratios learned in the previous unit stop being abstract triangle facts and start doing real measurement work. Every situation reduces to the same picture — a right triangle formed by an observer's line of sight, a horizontal distance, and a vertical height, linked by an angle of elevation or depression. CBSE keeps the scope deliberately narrow: at most two right triangles, angles restricted to $30^\circ$ , $45^\circ$ , $60^\circ$ , and no sine or cosine rule. Get the triangle-drawing and the tangent ratio right, and the rest is arithmetic with surds.

The core insight

Any height-or-distance word problem becomes a right triangle in which the angle of elevation or depression connects the unknown height (opposite side) to a known horizontal distance (adjacent side) through $\tan\theta = \text{opposite}/\text{adjacent}$ , and two-triangle problems are solved by writing one such equation per triangle and eliminating the shared unknown.

The core idea

Line of sight, elevation, depression. The line of sight is the straight line from the observer's eye to the object. When the object is above the observer's horizontal level, the angle between the line of sight and the horizontal is the angle of elevation (looking up). When the object is below, it is the angle of depression (looking down). Because the horizontal through the observer and the horizontal through the object are parallel lines cut by the same line of sight, alternate angles make the two equal: an angle of elevation of $\theta$ from the ground matches an angle of depression of $\theta$ from the top.

Turning the picture into a triangle. In every problem, the height (tower, cliff, building) forms one leg of a right triangle, the horizontal distance forms the other leg, and the angle of elevation or depression sits at the observer's vertex, opposite the height. Once the triangle is drawn and labelled, only two ratios are ever needed:

\tan\theta = \frac{\text{height (opposite)}}{\text{horizontal distance (adjacent)}}

where: $\theta$ is the angle of elevation or depression, the "opposite" side is the vertical height, and the "adjacent" side is the horizontal distance from the base to the observer. When the line of sight itself (the hypotenuse) is given or wanted, use instead

\sin\theta = \frac{\text{height}}{\text{line of sight}}, \qquad \cos\theta = \frac{\text{horizontal distance}}{\text{line of sight}}.

where: the line of sight is the hypotenuse of the right triangle, always the longest side, connecting the observer's eye directly to the object.

The standard values. Every numeric problem in this chapter uses only the five bookmark angles, so the ratio table below is worth knowing by heart rather than deriving each time:

$\theta$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
$\sin\theta$	$0$	$\tfrac12$	$\tfrac{1}{\sqrt2}$	$\tfrac{\sqrt3}{2}$	$1$
$\cos\theta$	$1$	$\tfrac{\sqrt3}{2}$	$\tfrac{1}{\sqrt2}$	$\tfrac12$	$0$
$\tan\theta$	$0$	$\tfrac{1}{\sqrt3}$	$1$	$\sqrt3$	undefined

Figure

A right triangle showing a tower as the vertical side, the ground distance as the horizontal side, the line of sight as the hypotenuse, and the angle of elevation marked at the observer

Two triangles, one unknown. Some problems give two observation points or two angles for the same object. The fix is always the same: name the unknown height $h$ , write a $\tan\theta$ equation for each right triangle, and combine the two equations — usually by division, so that $h$ cancels and only the horizontal distances remain, or by direct substitution.

Derivation ADistance between two points from two angles of elevation

The goal is to derive a general formula for the horizontal distance between two observation points on the same side of a tower, given the tower's height and the two different angles of elevation measured from those points, since this two-triangle elimination pattern is the one CBSE tests most often and every specific "ship" or "car" problem in this chapter is just this derivation with numbers substituted in.

Step 1. Set up the picture. Let $AB = h$ be the vertical tower, with $B$ its foot. Let $C$ and $D$ be two points on the horizontal ground through $B$ , with $D$ closer to the tower than $C$ , so that $BD < BC$ . Let the angle of elevation of the top $A$ from $D$ be $\beta$ and from $C$ be $\alpha$ , with $\alpha < \beta$ since $C$ is farther away and therefore sees a smaller angle.

Step 2. Write the tangent equation for the nearer triangle $ABD$ , right-angled at $B$ . Since the opposite side is $h$ and the adjacent side is $BD$ , $\tan\beta = h/BD$ , so $BD = h/\tan\beta$ .

Step 3. Write the tangent equation for the farther triangle $ABC$ , also right-angled at $B$ , with adjacent side $BC$ : $\tan\alpha = h/BC$ , so $BC = h/\tan\alpha$ .

Step 4. Subtract to isolate the distance between the two observation points, $CD = BC - BD$ , since $D$ lies between $B$ and $C$ . Substituting the two expressions from Steps 2 and 3 gives $CD = \dfrac{h}{\tan\alpha} - \dfrac{h}{\tan\beta} = h\left(\dfrac{1}{\tan\alpha} - \dfrac{1}{\tan\beta}\right)$ , which can be rearranged as $CD = \dfrac{h(\tan\beta - \tan\alpha)}{\tan\alpha\,\tan\beta}$ . Either the height $h$ or the distance $CD$ can now be found once the other three quantities ( $\alpha$ , $\beta$ , and one of $h$ or $CD$ ) are known, by solving this single equation for whichever quantity is missing.

Worked example

A boy standing $20\ \text{m}$ from the foot of a tower observes the top of the tower at an angle of elevation of $60^\circ$ . Find the height of the tower. (Take $\sqrt3 = 1.73$ .)

Step 1. Draw the triangle. Let the tower be $AB = h$ and the boy's position be $C$ , with $BC = 20\ \text{m}$ the horizontal distance and $\angle ACB = 60^\circ$ the angle of elevation, right-angled at $B$ .

Step 2. Apply the tangent ratio: $\tan 60^\circ = AB/BC = h/20$ .

Step 3. Substitute $\tan 60^\circ = \sqrt3$ and solve: $\sqrt3 = h/20 \implies h = 20\sqrt3 = 20 \times 1.73 = 34.6\ \text{m}$ . The height of the tower is 34.6 m.

Competitive stretch

From the top of a $50\ \text{m}$ lighthouse, the angle of depression of an approaching ship changes from $30^\circ$ to $60^\circ$ . Find the distance travelled. (Take $\sqrt3 = 1.73$ .)

Step 1. By alternate angles, the angles of elevation from the ship's two positions $C$ (farther, angle $30^\circ$ ) and $D$ (nearer, angle $60^\circ$ ) equal the angles of depression from the top $A$ , with $AB = 50\ \text{m}$ .

Step 2. Using the derivation above with $\alpha = 30^\circ$ , $\beta = 60^\circ$ , $h = 50$ : $BD = 50/\tan 60^\circ = 50/\sqrt3$ and $BC = 50/\tan 30^\circ = 50\sqrt3$ .

Step 3. The distance travelled is $CD = BC - BD = 50\sqrt3 - 50/\sqrt3 = 100/\sqrt3 = 100\sqrt3/3 \approx 57.67\ \text{m}$ , using $\sqrt3 = 1.73$ . The ship travels approximately 57.67 m towards the lighthouse.

Common traps

Mixing up elevation and depression. The angle of depression is measured from the observer's horizontal downward to the object — draw it at the observer's own vertex, not at the ground.
Wrong side labels. Confusing "opposite" and "adjacent" relative to the marked angle flips the ratio upside down, effectively using $\cot\theta$ instead of $\tan\theta$ .
Forgetting the observer's eye height. If a problem states the observer's height above the ground, that value must be added to the triangle's computed answer to get the object's true height.
Leaving surds unrationalised. $\dfrac{100}{\sqrt3}$ is numerically fine, but CBSE marking schemes expect it rationalised to $\dfrac{100\sqrt3}{3}$ before decimals are substituted.

Ask the tutor: Stuck setting up the right triangle for a two-point or lighthouse-and-ship problem, or unsure whether to use sin, cos, or tan? Ask the AI tutor to sketch the triangle with you, mark the angles correctly, and check your equation before you solve.

How do I know whether to use tan, sin, or cos in a given problem?

Look at which two sides the problem connects to the angle. If it links the height (opposite) to the horizontal ground distance (adjacent), use $\tan\theta$ — this covers most CBSE problems. If the line of sight itself (the slanted distance, the hypotenuse) is given or asked for, use $\sin\theta$ (height and hypotenuse) or $\cos\theta$ (horizontal distance and hypotenuse) instead. Drawing and labelling the triangle first makes this choice automatic rather than guesswork.

Why does the angle of depression equal the angle of elevation from the other end?

Because the observer's horizontal and the object's horizontal are parallel lines, and the line of sight is a single straight line cutting across both — a transversal. The angle the line of sight makes with the upper horizontal (depression, measured downward) and the angle it makes with the lower horizontal (elevation, measured upward) are alternate angles between those parallel lines, and alternate angles are always equal. This is why lighthouse-and-ship problems can be redrawn entirely from the ship's point of view using elevation angles.

What do I do if a problem gives two different angles from two different points, but I only need one equation?

You still need both angles, but only one of them may appear explicitly in the final step. Set up a $\tan\theta$ equation for each observation point separately — that gives two equations in two unknowns (usually the height $h$ and one distance). Dividing one equation by the other, or substituting one into the other, eliminates the shared unknown and leaves a single equation you can solve directly, exactly as in the derivation above.

Heights and distances

The core idea

Worked example

Competitive stretch

Common traps

Before this