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Some Applications of Trigonometry - Heights and Distances

Grade 10CBSE

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

🔑Concepts

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Line of Sight: This is the straight line drawn from the eye of an observer to the point in the object being viewed. Visually, it acts as the hypotenuse of the right-angled triangle formed between the observer and the object.

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Angle of Elevation: When an observer looks at an object situated above their horizontal eye level, the angle formed between the horizontal line and the line of sight is the angle of elevation. Visually, imagine looking up from the ground to the top of a flagpole; the angle between the ground and your line of sight is the elevation.

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Angle of Depression: When an observer looks down at an object situated below their horizontal eye level, the angle formed between the horizontal line and the line of sight is the angle of depression. Visually, if you are standing on a balcony looking down at a car on the road, the angle between your straight-ahead horizontal gaze and your downward look is the depression.

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Horizontal Level: This is the reference line passing through the observer's eyes, parallel to the ground. In geometric diagrams, this is usually the base of the triangle or a line parallel to the base.

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Equality of Elevation and Depression: The angle of depression from a point AA to a point BB is numerically equal to the angle of elevation of AA as seen from BB. Visually, this is because the horizontal lines at the top and bottom are parallel, making these alternate interior angles.

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Trigonometric Ratio Selection: To solve height and distance problems, we choose a trigonometric ratio that relates the known side and the unknown side. Usually, tan⁡θ\tan \theta is used when dealing with the height (opposite) and the distance (adjacent).

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Right Triangle Representation: Real-world objects like buildings, towers, and trees are assumed to be vertical and perpendicular to the horizontal ground. This allows us to model these problems using right-angled triangles where the vertical object is the perpendicular and the ground distance is the base.

📐Formulae

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

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

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

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

tan⁡45∘=1\tan 45^{\circ} = 1

tan⁡60∘=3\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:

Let ABAB be the tower of height hh and CC be the point on the ground. Given: Distance BC=15BC = 15 m and ∠ACB=60∘\angle ACB = 60^{\circ}. In right △ABC\triangle ABC: tan⁡60∘=ABBC\tan 60^{\circ} = \frac{AB}{BC} 3=h15\sqrt{3} = \frac{h}{15} h=153 mh = 15\sqrt{3} \text{ m} If we take 3≈1.732\sqrt{3} \approx 1.732, then h=15×1.732=25.98h = 15 \times 1.732 = 25.98 m.

Explanation:

The problem provides the adjacent side (distance from the foot) and asks for the opposite side (height). Therefore, the tangent ratio is the most direct formula to use.

Problem 2:

From the top of a 77 m high building, the angle of elevation of the top of a cable tower is 60∘60^{\circ} and the angle of depression of its foot is 45∘45^{\circ}. Determine the height of the tower.

Solution:

Let AB=7AB = 7 m be the building and CDCD be the tower. Let AEAE be the horizontal line from the top of the building to the tower. Then AB=EC=7AB = EC = 7 m. In right △ABC\triangle ABC, the angle of depression is 45∘45^{\circ}, so ∠ACB=45∘\angle ACB = 45^{\circ} (alternate angles). tan⁡45∘=ABBC⇒1=7BC⇒BC=7 m\tan 45^{\circ} = \frac{AB}{BC} \Rightarrow 1 = \frac{7}{BC} \Rightarrow BC = 7 \text{ m} Since AE=BCAE = BC, AE=7AE = 7 m. In right △AED\triangle AED: tan⁡60∘=DEAE⇒3=DE7⇒DE=73 m\tan 60^{\circ} = \frac{DE}{AE} \Rightarrow \sqrt{3} = \frac{DE}{7} \Rightarrow DE = 7\sqrt{3} \text{ m} Total height of tower CD=CE+DE=7+73=7(1+3) mCD = CE + DE = 7 + 7\sqrt{3} = 7(1 + \sqrt{3}) \text{ m}.

Explanation:

This solution breaks the tower into two parts: the portion equal to the building height and the portion above it. We use the angle of depression to find the distance between the two structures, which then serves as the base for the second triangle to find the remaining height.