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Trigonometry - Heights and Distances

Grade 10ICSE

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

🔑Concepts

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Line of Sight: The imaginary straight line joining the eye of an observer to the point in the object being viewed. In a geometric diagram, this represents the hypotenuse of the right-angled triangle formed between the observer and the object.

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Horizontal Level: A line drawn parallel to the surface of the earth starting from the eye of the observer. It acts as the reference base line (0 degrees) from which all angles of elevation or depression are measured.

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Angle of Elevation: When the object being viewed is above the horizontal level, the angle formed between the horizontal line and the line of sight is called the angle of elevation. Visually, this is the angle created when the observer tilts their head 'up' to see the target.

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Angle of Depression: When the object being viewed is below the horizontal level, the angle formed between the horizontal line and the line of sight is called the angle of depression. Visually, this is the angle created when the observer looks 'down' toward the target.

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Equality of Elevation and Depression: The angle of depression of a point PP as seen from a point OO is equal to the angle of elevation of point OO as seen from point PP. This is because the horizontal lines at both points are parallel, making the two angles alternate interior angles in a transversal system.

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Right-Angled Triangle Modeling: Heights and distances problems are solved by representing the scenario as a right-angled triangle. The vertical object (like a building or tower) is the perpendicular, the ground distance is the base, and the line of sight is the hypotenuse.

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Trigonometric Application: The tangent ratio is the most frequently used tool in these problems because it links the height (opposite side) and the distance (adjacent side) without needing the length of the line of sight (hypotenuse).

📐Formulae

tan⁡θ=PerpendicularBase\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}

sin⁡θ=PerpendicularHypotenuse\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}

cos⁡θ=BaseHypotenuse\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}}

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}

3≈1.732\sqrt{3} \approx 1.732

2≈1.414\sqrt{2} \approx 1.414

💡Examples

Problem 1:

The angle of elevation of the top of a tower from a point on the ground, which is 3030 m away from the foot of the tower, is 30∘30^{\circ}. Find the height of the tower.

Solution:

Let hh be the height of the tower ABAB and dd be the distance from the point CC to the foot of the tower BB. \n\n Given: \n Distance BC=30BC = 30 m \n Angle of elevation ∠ACB=30∘\angle ACB = 30^{\circ} \n\n In right-angled △ABC\triangle ABC: \n tan⁡30∘=ABBC\tan 30^{\circ} = \frac{AB}{BC} \n 13=h30\frac{1}{\sqrt{3}} = \frac{h}{30} \n h=303h = \frac{30}{\sqrt{3}} \n h=30×33×3h = \frac{30 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \n h=3033=103h = \frac{30\sqrt{3}}{3} = 10\sqrt{3} m \n h=10×1.732=17.32h = 10 \times 1.732 = 17.32 m

Explanation:

We use the tangent ratio because we are given the base (distance from tower) and need to find the perpendicular (height of the tower).

Problem 2:

From the top of a 7575 m high lighthouse, the angles of depression of two ships are 30∘30^{\circ} and 45∘45^{\circ}. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Solution:

Let ABAB be the lighthouse of height 7575 m. Let CC and DD be the positions of the two ships. \n\n In △ABC\triangle ABC (closer ship with 45∘45^{\circ} angle): \n tan⁡45∘=ABBC\tan 45^{\circ} = \frac{AB}{BC} \n 1=75BC⇒BC=751 = \frac{75}{BC} \Rightarrow BC = 75 m \n\n In △ABD\triangle ABD (farther ship with 30∘30^{\circ} angle): \n tan⁡30∘=ABBD\tan 30^{\circ} = \frac{AB}{BD} \n 13=75BD⇒BD=753\frac{1}{\sqrt{3}} = \frac{75}{BD} \Rightarrow BD = 75\sqrt{3} m \n\n Distance between ships CD=BD−BCCD = BD - BC: \n CD=753−75CD = 75\sqrt{3} - 75 \n CD=75(3−1)CD = 75(\sqrt{3} - 1) \n CD=75(1.732−1)=75(0.732)=54.9CD = 75(1.732 - 1) = 75(0.732) = 54.9 m

Explanation:

The distance between the ships is the difference between their respective horizontal distances from the foot of the lighthouse. We solve for both distances using the tangent ratio and subtract them.