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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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The line of sight is an imaginary line drawn from the eye of an observer to the object being viewed. The angle formed between the line of sight and the horizontal level is either the angle of elevation or the angle of depression.

Diagram showing the line of sight and horizontal line from an observer's eye.
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Angle of Elevation: When the object being viewed is above the horizontal level, the angle between the horizontal line and the line of sight is called the angle of elevation. In most problems, the ground is considered the horizontal level.

Right-angled triangle showing the angle of elevation theta.
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Angle of Depression: When the object being viewed is below the horizontal level, the angle between the horizontal line through the observer's eye and the line of sight is called the angle of depression.

Diagram showing angle of depression from a high point to a low point.
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Alternate Angles: In problems involving the angle of depression, the angle of depression from the top of an object is equal to the angle of elevation of the top from the object on the ground because the horizontal lines are parallel.

Parallel lines showing alternate interior angles are equal.

📐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.

Problem 3:

An observer 1.5 m1.5\ m tall is 28.5 m28.5\ m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45∘45^\circ. What is the height of the chimney?

Geometry diagram of observer looking at chimney with height 1.5m and 45 degree angle.

Solution:

Let ABAB be the chimney and CDCD be the observer. Given: CD=1.5 mCD = 1.5\ m and distance BD=28.5 mBD = 28.5\ m. Draw CECE parallel to BDBD, meeting ABAB at EE. Then, CE=BD=28.5 mCE = BD = 28.5\ m and EB=CD=1.5 mEB = CD = 1.5\ m. In right-angled triangle △ACE\triangle ACE, we have: tan⁡45∘=AECE\tan 45^\circ = \frac{AE}{CE} 1=AE28.51 = \frac{AE}{28.5} AE=28.5 mAE = 28.5\ m Total height of chimney AB=AE+EBAB = AE + EB AB=28.5+1.5=30 mAB = 28.5 + 1.5 = 30\ m The height of the chimney is 30 m30\ m.

Explanation:

To solve this, we model the observer and chimney as vertical lines. We create a right triangle above the observer's eye level and use the tangent ratio since we know the horizontal distance and the angle.

Problem 4:

The shadow of a tower standing on a level ground is found to be 40 m40\ m longer when the Sun's altitude is 30∘30^\circ than when it is 60∘60^\circ. Find the height of the tower.

Diagram of a tower showing two shadow lengths for 30 and 60 degree solar altitudes.

Solution:

Let hh be the height of the tower ABAB. Let BC=xBC = x be the shadow when the angle is 60∘60^\circ. Then the shadow BD=x+40BD = x + 40 when the angle is 30∘30^\circ. In △ABC\triangle ABC: tan⁡60∘=hx⇒3=hx⇒x=h3\tan 60^\circ = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}} In △ABD\triangle ABD: tan⁡30∘=hx+40\tan 30^\circ = \frac{h}{x + 40} 13=hx+40\frac{1}{\sqrt{3}} = \frac{h}{x + 40} x+40=h3x + 40 = h\sqrt{3} Substituting x=h3x = \frac{h}{\sqrt{3}}: h3+40=h3\frac{h}{\sqrt{3}} + 40 = h\sqrt{3} h+403=3hh + 40\sqrt{3} = 3h 2h=4032h = 40\sqrt{3} h=203≈34.64 mh = 20\sqrt{3} \approx 34.64\ m The height of the tower is 203 m20\sqrt{3}\ m.

Explanation:

This problem involves two different positions of the sun. We set up two trigonometric equations based on the two angles of elevation and solve for the common height 'h' by substituting the variable representing the shorter shadow.