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Trigonometry - Simple 2-D problems on Heights and Distances

Grade 9ICSE

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

🔑Concepts

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Line of Sight: This is the imaginary straight line drawn from the eye of an observer to the point in the object viewed by the observer. 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: The horizontal level is the straight line originating from the observer's eye parallel to the ground. It serves as the reference line (the base or a line parallel to the base) from which 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 line of sight and the horizontal level is called the angle of elevation. Visually, imagine looking up from the ground to the top of a flag pole; the angle is formed at your feet between the ground and your line of sight.

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Angle of Depression: When the object being viewed is below the horizontal level, the angle formed between the line of sight and the horizontal level is called the angle of depression. Visually, if you are standing on a balcony looking down at a car, the angle is formed between your straight-ahead gaze and the downward diagonal path to the car.

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Equality of Elevation and Depression: By the geometric property of alternate interior angles, the angle of depression from a point AA to a point BB is always equal to the angle of elevation from point BB to point AA. This allows us to represent the angle of depression inside the triangle at the ground level for easier calculation.

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Right-Angled Triangle Representation: Simple 2-D height and distance problems are modeled using a right-angled triangle. The vertical side (ABAB) typically represents the height of an object like a tower or building, the horizontal side (BCBC) represents the distance from the foot, and the hypotenuse (ACAC) represents the line of sight.

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Trigonometric Ratio Selection: To solve these problems, we select ratios based on known and unknown sides. Most frequently, tan⁡θ\tan \theta is used because it relates the height (opposite) and the distance (adjacent), but sin⁡θ\sin \theta or cos⁡θ\cos \theta are used if the length of the line of sight (hypotenuse) is involved.

📐Formulae

sin⁡θ=Opposite Side (Perpendicular)Hypotenuse\sin \theta = \frac{\text{Opposite Side (Perpendicular)}}{\text{Hypotenuse}}

cos⁡θ=Adjacent Side (Base)Hypotenuse\cos \theta = \frac{\text{Adjacent Side (Base)}}{\text{Hypotenuse}}

tan⁡θ=Opposite Side (Perpendicular)Adjacent Side (Base)\tan \theta = \frac{\text{Opposite Side (Perpendicular)}}{\text{Adjacent Side (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}

Height=Distance×tan⁡(Angle of Elevation)\text{Height} = \text{Distance} \times \tan(\text{Angle of Elevation})

💡Examples

Problem 1:

A ladder is placed against a wall such that it reaches the top of the wall of height 6 m6\ m. If the ladder makes an angle of 60∘60^\circ with the ground, find the length of the ladder.

Solution:

  1. Let ABAB be the wall of height 6 m6\ m and ACAC be the length of the ladder.
  2. In the right-angled △ABC\triangle ABC, the angle of elevation ∠ACB=60∘\angle ACB = 60^\circ.
  3. We need to find the hypotenuse (ACAC) and we know the perpendicular (AB=6 mAB = 6\ m).
  4. Using the sine ratio: sin⁡60∘=ABAC\sin 60^\circ = \frac{AB}{AC}
  5. 32=6AC\frac{\sqrt{3}}{2} = \frac{6}{AC}
  6. AC=6×23=123AC = \frac{6 \times 2}{\sqrt{3}} = \frac{12}{\sqrt{3}}
  7. Rationalizing the denominator: AC=1233=43 mAC = \frac{12\sqrt{3}}{3} = 4\sqrt{3}\ m
  8. Taking 3≈1.732\sqrt{3} \approx 1.732, AC=4×1.732=6.928 mAC = 4 \times 1.732 = 6.928\ m.

Explanation:

In this problem, we identify the ladder as the hypotenuse of a right-angled triangle. Since we are given the height of the wall (opposite side to the angle) and need the hypotenuse, we apply the sine ratio.

Problem 2:

From the top of a tower 50 m50\ m high, the angle of depression of a ball on the ground is 30∘30^\circ. Find the distance of the ball from the foot of the tower.

Solution:

  1. Let PQPQ be the tower of height 50 m50\ m and RR be the position of the ball.
  2. The angle of depression is given as 30∘30^\circ. This is equal to the angle of elevation ∠PRQ=30∘\angle PRQ = 30^\circ due to alternate angles.
  3. In right-angled △PQR\triangle PQR, we know the perpendicular (PQ=50 mPQ = 50\ m) and need to find the base (QRQR).
  4. Using the tangent ratio: tan⁡30∘=PQQR\tan 30^\circ = \frac{PQ}{QR}
  5. 13=50QR\frac{1}{\sqrt{3}} = \frac{50}{QR}
  6. QR=503 mQR = 50\sqrt{3}\ m
  7. QR=50×1.732=86.6 mQR = 50 \times 1.732 = 86.6\ m.

Explanation:

The angle of depression is measured from the horizontal line at the top of the tower. We translate this to the interior angle at the ground level using the property of parallel lines. We then use the tangent ratio to find the horizontal distance.