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Some Applications of Trigonometry - Angle of Elevation and Angle of Depression

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Line of Sight: When an observer looks at an object, the imaginary straight line drawn from the eye of the observer to the point in the object being viewed is called the line of sight. Visually, this is the hypotenuse of the right-angled triangle formed between the observer and the object.

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Horizontal Level: This is the straight horizontal line passing through the observer's eye, parallel to the ground. In a geometric diagram, this represents the base or a line parallel to the base from which angles are measured.

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Angle of Elevation: The angle formed by the line of sight with the horizontal level when the object is above the horizontal level. To visualize this, imagine looking up at the top of a tower from the ground; the angle between your straight-ahead gaze and your upward gaze is the angle of elevation.

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Angle of Depression: The angle formed by the line of sight with the horizontal level when the point being viewed is below the horizontal level. Visually, if you are standing on a balcony looking down at a car, the angle between the horizontal line at your eye level and the line pointing down to the car is the angle of depression.

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Relationship Between Elevation and Depression: For two points A (higher) and B (lower), the angle of depression of B as seen from A is equal to the angle of elevation of A as seen from B. This is because these two angles form alternate interior angles between the horizontal line at A and the ground level (or horizontal line at B).

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Right-Angled Triangle Modeling: Real-world scenarios like heights of buildings, distances of ships from lighthouses, or lengths of shadows are modeled as right-angled triangles. The vertical height (tower/building) is the perpendicular, the horizontal distance is the base, and the line of sight is the hypotenuse.

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Choosing the Right Ratio: To find a missing height or distance, identify the known and unknown sides relative to the given angle. Usually, tan⁡θ\tan \theta (Perpendicular/Base) is used for heights and distances, while sin⁡θ\sin \theta (Perpendicular/Hypotenuse) is used for lengths like kite strings or ladders.

📐Formulae

sin⁡θ=Side opposite to angle θHypotenuse=PerpendicularHypotenuse\sin \theta = \frac{\text{Side opposite to angle } \theta}{\text{Hypotenuse}} = \frac{\text{Perpendicular}}{\text{Hypotenuse}}

cos⁡θ=Side adjacent to angle θHypotenuse=BaseHypotenuse\cos \theta = \frac{\text{Side adjacent to angle } \theta}{\text{Hypotenuse}} = \frac{\text{Base}}{\text{Hypotenuse}}

tan⁡θ=Side opposite to angle θSide adjacent to angle θ=PerpendicularBase\tan \theta = \frac{\text{Side opposite to angle } \theta}{\text{Side adjacent to angle } \theta} = \frac{\text{Perpendicular}}{\text{Base}}

tan⁡30∘=13,tan⁡45∘=1,tan⁡60∘=3\tan 30^{\circ} = \frac{1}{\sqrt{3}}, \tan 45^{\circ} = 1, \tan 60^{\circ} = \sqrt{3}

sin⁡30∘=12,sin⁡45∘=12,sin⁡60∘=32\sin 30^{\circ} = \frac{1}{2}, \sin 45^{\circ} = \frac{1}{\sqrt{2}}, \sin 60^{\circ} = \frac{\sqrt{3}}{2}

cos⁡30∘=32,cos⁡45∘=12,cos⁡60∘=12\cos 30^{\circ} = \frac{\sqrt{3}}{2}, \cos 45^{\circ} = \frac{1}{\sqrt{2}}, \cos 60^{\circ} = \frac{1}{2}

💡Examples

Problem 1:

A tower stands vertically on the ground. From a point on the ground, which is 1515 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60∘60^{\circ}. Find the height of the tower.

Solution:

Step 1: Let ABAB be the height of the tower (hh) and BCBC be the distance from the foot of the tower to the point of observation, so BC=15BC = 15 m. The angle of elevation ∠ACB=60∘\angle ACB = 60^{\circ}. Step 2: In the right-angled triangle ABCABC, we use the tangent ratio because we have the base and need the perpendicular. tan⁡60∘=ABBC\tan 60^{\circ} = \frac{AB}{BC} 3=h15\sqrt{3} = \frac{h}{15} Step 3: Solve for hh: h=153h = 15\sqrt{3} Therefore, the height of the tower is 15315\sqrt{3} m (approx 25.9825.98 m).

Explanation:

We identify the height as the 'Perpendicular' and the distance from the foot as the 'Base'. The tangent ratio connects these two values with the given angle of elevation.

Problem 2:

The angle of depression of a car parked on the road from the top of a 7575 m high tower is 30∘30^{\circ}. Find the distance of the car from the base of the tower.

Solution:

Step 1: Let PQPQ be the tower of height 7575 m. Let RR be the position of the car. The angle of depression is 30∘30^{\circ}. Step 2: The angle of depression from the top of the tower to the car is equal to the angle of elevation of the top of the tower from the car (alternate interior angles). So, ∠PRQ=30∘\angle PRQ = 30^{\circ}. Step 3: In △PQR\triangle PQR, we use tan⁡30∘\tan 30^{\circ}: tan⁡30∘=PQQR\tan 30^{\circ} = \frac{PQ}{QR} 13=75QR\frac{1}{\sqrt{3}} = \frac{75}{QR} Step 4: Solve for QRQR: QR=753QR = 75\sqrt{3} Therefore, the car is 75375\sqrt{3} m away from the base of the tower.

Explanation:

This problem uses the property that the angle of depression equals the angle of elevation at the target point. We then use the tangent ratio to find the horizontal distance (Base).