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Question
A vertical flagstaff stands on the top of a building. From a point on the ground 24 m from the base of the building, the angles of elevation of the top and bottom of the flagstaff are 60∘60^\circ and 45∘45^\circ respectively. The height of the flagstaff is:
(A)24(3+1)24(\sqrt{3} + 1) m
(B)24(3−1)24(\sqrt{3} - 1) m
(C)24324\sqrt{3} m
(D)2424 m
Solution Path
Two right triangles with common base 24 m. Use tan⁡45∘\tan 45^\circ for building height (24 m), tan⁡60∘\tan 60^\circ for total height (24324\sqrt{3} m). Flagstaff = 24(3−1)24(\sqrt{3} - 1) m.
01Question Setup
1/4
A flagstaff stands on a building. From 24 m away, angles of elevation to top and bottom of flagstaff are 60∘60^\circ and 45∘45^\circ. Find the flagstaff height.
Two angles, one distance, two unknowns
02Triangle Diagram
2/4
Draw two right triangles sharing the same base (24 m). The 45∘45^\circ angle gives the building height, and the 60∘60^\circ angle gives the total height (building + flagstaff).
tan⁡45∘=h124\tan 45^\circ = \frac{h_1}{24} and tan⁡60∘=h1+h224\tan 60^\circ = \frac{h_1 + h_2}{24}
03Solve Both TrianglesKEY INSIGHT
3/4
From tan⁡45∘=1\tan 45^\circ = 1, building height h1=24h_1 = 24 m. From tan⁡60∘=3\tan 60^\circ = \sqrt{3}, total height =243= 24\sqrt{3}. Subtract to get flagstaff height.
h2=243−24=24(3−1)h_2 = 24\sqrt{3} - 24 = 24(\sqrt{3} - 1) m
04Final Answer
4/4
The flagstaff height is 24(3−1)≈17.524(\sqrt{3} - 1) \approx 17.5 m.
Answer (B):24(3−1) m\boxed{\text{Answer (B)}: 24(\sqrt{3} - 1) \text{ m}}
Concepts from this question2 concepts unlocked
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Elevation and Depression Triangle Setup

EASY

The angle of elevation is the angle between the horizontal and the line of sight when looking upward. The angle of depression is the angle when looking downward. Both create right triangles where tan(angle) = opposite/adjacent is the primary relation.

tan⁡(α)=heightdistance,Elevation angle=Depression angle (alternate angles)\tan(\alpha) = \frac{\text{height}}{\text{distance}}, \quad \text{Elevation angle} = \text{Depression angle (alternate angles)}

Every heights and distances problem starts with identifying whether you are looking up or down. Getting this wrong means the entire triangle is set up incorrectly. The alternate angle property is the single most used fact in this chapter.

Basic height problemsDepression distance problemsTower height from ground
Practice (12 Qs) →

Two-Point Observation Method

STANDARD

When an object is observed from two different points at known separation, two right triangles share the same height. Setting up tan equations for both angles and eliminating the unknown distance gives the height directly.

h=d⋅tan⁡α⋅tan⁡βtan⁡β−tan⁡α(same side, β>α)h = \frac{d \cdot \tan\alpha \cdot \tan\beta}{\tan\beta - \tan\alpha} \quad (\text{same side, } \beta > \alpha)

This is the most frequently tested pattern in heights and distances. JEE problems love giving two angles from two points and asking for the height. Memorizing this formula saves 2-3 minutes per problem.

Two-angle height problemsWalking toward tower problemsDistance between two points
Practice (9 Qs) →
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