Heights and Distances: Turn a View into a Triangle

Start with the horizontal line

Imagine looking from a point at the top of a tower. The straight segment joining your eye to the target is the line of sight. Through your eye, draw a horizontal reference line. The angle is measured from that horizontal—not automatically from the tower or from the vertical.

When the target is above the horizontal, the angle is an angle of elevation (उन्नयन कोण). When the target is below it, the angle is an angle of depression (अवनमन कोण). Both descriptions refer to the observer’s viewpoint.

Suppose a point on level ground is viewed from a tower top at a depression angle of 30°. The horizontal through the tower top is parallel to the horizontal ground. The line of sight cuts both. Alternate interior angles show that the elevation of the tower top from the ground point is also 30°.

The two angles are equal because the horizontals are parallel. They are not vertically opposite angles, and they are not measured at the same vertex. Marking both horizontal lines makes the reason visible.

If a line of sight makes 30° with the horizontal, it makes 60° with a vertical line at that same point. Confusing these two angles changes the ratio you use.

Before solving, ask: who is looking, where is that observer’s horizontal, and is the target above or below it? A clear answer to these three questions is often the most important part of the problem. Add the right-angle mark only where a stated vertical meets a horizontal.

Choose a triangle before choosing a formula

A word problem gives a scene; your sketch turns it into mathematics. Draw a vertical object meeting horizontal ground. Join the observer to the target. Mark the known angle at the observer, then label the lengths as opposite, adjacent or hypotenuse relative to that angle.

A horizontal distance to a tower’s foot is usually the adjacent leg. The height above the observer’s horizontal is the opposite leg. The sloping line of sight is the hypotenuse. Those three distances are different, even if a small sketch makes two of them look similar.

Choose a ratio containing the known length and the required length:

If a model line of sight is 10 m long and makes 30° with a horizontal baseline, the rise is $10\sin30^\circ=5$m. Its horizontal span is $10\cos30^\circ=5\sqrt3$m. The sloping 10 m must not be placed in tangent’s denominator as though it were a horizontal distance.

Record the assumptions represented by the sketch: the object is vertical, the chosen ground segment is horizontal, and the named points lie in the same vertical plane. These are geometric details that tell you where the right triangle exists.

Use a four-line solution: label the sketch, write the ratio equation, substitute exact values, and give the answer with units. After solving, compare the result with the drawing’s geometry. The sloping side must remain the longest, and an angle of elevation is always measured from the horizontal reference you have actually drawn.

Find a height with one right triangle

Begin with an observer represented by a point on horizontal ground. Let $d$ be the horizontal distance to the foot of a vertical object, and let $h$ be its height. If the elevation of its top is $\theta$, the model gives:

For a tower 12 m from the observation point and an elevation of 60°, $h=12\sqrt3$m. This is an exact answer. If an approximate height is needed, state the approximation you use for $\sqrt3$ and round only at the end.

The equation also works backward. If a vertical object has height 18 m and its elevation is 45°, then $\tan45^\circ=18/d$. Because tangent 45° is 1, the horizontal distance is 18 m.

The three standard angles give a useful pattern. For a fixed horizontal distance, a larger elevation produces a greater height: tangent 30° is $1/\sqrt3$, tangent 45° is 1, and tangent 60° is $\sqrt3$. For a fixed height, walking farther away makes the elevation smaller.

Explore this pattern in the model. Use 30°, 45° and 60° to practise the exact-value problems on this page. If the model also allows other angles, use them to explore the changing shape; the practice questions keep the familiar standard-angle route.

Finish every answer by saying which distance you found. “18 m” by itself could mean a height, a horizontal distance or a sloping length. Naming the quantity helps you check that the equation answered the question rather than a different part of the diagram.

Include eye height—and read depression carefully

An observer’s eye is often above the ground. In that case, the opposite leg of the sight triangle is the height above eye level, not the complete object height.

Let the eye height be $e$, the horizontal distance to the object’s foot be $d$, and its total height be $h$. If both feet are on the same horizontal level, then:

For an eye height of 1.6 m, a horizontal distance of 15 m and an elevation of 45°, the height above the eye is 15 m. The total object height is $1.6+15=16.6$m. Ignoring the eye height would answer a different question.

Now reverse the view. An observer at a point 20 m above horizontal ground looks down at a ground point with a depression angle of 30°. Transfer that angle to the lower horizontal using parallel lines. The ground-level triangle has opposite leg 20 m, so:

Here 20 m is the vertical separation between the two sight points. Do not add a person’s height again if the given 20 m already measures the observation point’s height above the ground.

Read what each height measures. A tower’s structural height, a viewing platform’s level and a person’s eye level are not interchangeable. Draw an extra horizontal line through the eye and mark the small vertical segment below it. That extra line is often all that is needed to prevent a missing-height error.

Two positions can reveal an unknown height

Sometimes neither the height nor the distance to the object is known. Two observations can give two equations. The important move is to use one height symbol in both right triangles.

Two observation points lie on the same straight horizontal line, on the same side of a vertical tower. The nearer point is 20 m closer to the tower than the farther point. Elevations of the top are 60° at the nearer point and 30° at the farther point. Both observations are modelled from ground level.

Let the nearer distance be $x$m and the tower height be $h$m. The farther distance is $x+20$m. Write one tangent equation for each triangle:

Because both expressions equal the same height, set them equal. Multiplying by $\sqrt3$ gives $3x=x+20$, so $x=10$. Therefore $h=10\sqrt3$m, and the farther distance is 30 m.

Check both views: $h/10=\sqrt3$, while $h/30=1/\sqrt3$. The larger angle belongs to the nearer observer, as the sketch suggests.

Draw the position order before writing distances. “On the same side” makes the farther distance $x+20$. If the points were on opposite sides of the tower, their separation would instead be the sum of their individual distances. The words change the equations. Keep the target height common, but give every distinct horizontal segment its correct meaning.

Separate a building from the flagstaff above it

Two right triangles can share a horizontal baseline but end at different heights. Draw both sight lines from the same observation point, and give the two vertical targets different labels.

A vertical building is 12 m tall. From a ground point on the same horizontal level as its foot, the building’s top has elevation 30°. A vertical flagstaff rises from that top; the elevation of the flagstaff’s top is 45°. Find the flagstaff’s length.

First use the lower sight triangle to find the common horizontal distance $d$:

Then use the upper sight triangle. Let $H$ be the combined height of building and flagstaff:

The flagstaff is only the extra part, so its length is:

This is the difference of two heights, not the difference of the two angles. Subtracting 45°−30° and treating 15° as an elevation from the ground point would create a different triangle and would not solve the stated problem.

Check what your answer represents. The 45° sight triangle ends at the flagstaff top, so its opposite leg includes both structures. The 30° sight triangle ends at the building top. Each triangle has one right angle and the same horizontal distance.

Bring the route together: choose the observer’s horizontal, label the requested angle, match each length to a side, solve with exact standard values, and return to the named quantity. These steps now cover the one- and two-triangle height problems in your Class 10 route.