JEE Maths›Heights & Distances›Visual Solution

Visual SolutionTricky

Question

A man on the deck of a ship, 12 m above water level, observes that the angle of elevation of the top of a cliff is

60^\circ

and the angle of depression of the base of the cliff is

30^\circ

. The height of the cliff is:

(A)

36

(B)

48

(C)

24\sqrt{3}

(D)

60

Solution Path

Depression to base gives

d = 12\sqrt{3}

m. Elevation to top gives height above deck = 36 m. Total cliff = 36 + 12 = 48 m.

01Question Setup

1/4

A man on a ship 12 m above water sees a cliff. Elevation to cliff top is

60^\circ

, depression to cliff base is

30^\circ

. Find the cliff height.

Combined elevation and depression from the same point

02Ship and Cliff Diagram

2/4

The observer is 12 m above water. Depression to cliff base gives horizontal distance

d

. Elevation to cliff top gives the height above deck level

h_1

\tan 30^\circ = \frac{12}{d}

and

\tan 60^\circ = \frac{h_1}{d}

03Two Right TrianglesKEY INSIGHT

3/4

From depression:

d = 12\sqrt{3}

m. From elevation:

h_1 = d\sqrt{3} = 36

m. Total cliff =

h_1 + 12 = 48

m. The common trap is forgetting to add the 12 m deck height.

Total cliff = 36 + 12 = 48 m

04Final Answer

4/4

The cliff height is 48 m. Sanity check: 12 (deck) + 36 (above deck) = 48 m.

\boxed{\text{Answer (B)}: 48 \text{ m}}

Concepts from this question2 concepts unlocked

★

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(\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) →

Moving Observer Technique

STANDARD

When an observer moves toward or away from an object, the angle of elevation changes. The distance moved and the two angles give enough information to find the height. This is equivalent to the two-point observation with the separation being the distance walked.

h = \frac{d \cdot \tan\alpha \cdot \tan\beta}{\tan\beta - \tan\alpha} \quad (d = \text{distance moved})

Moving observer problems are disguised two-point problems. Recognizing this saves time. The formula is identical; only the context changes.

Walking toward tower problemsShip approaching lighthouseCar approaching building

Practice (8 Qs) →

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