This MCQ module is based on: Heights and Distances – Introduction
TOPIC 23 OF 35
Heights and Distances – Introduction
🎓 Class 10
Mathematics
CBSE
Theory
Ch 9 — Some Applications of Trigonometry
⏱ ~30 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Heights and Distances – Introduction
Targeting Class 10 level in Trigonometry, with Intermediate difficulty.
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9.1 Heights and Distances
In Chapter 8 you studied how trigonometric ratios relate the sides of a right triangle to its acute angles. This chapter turns those ratios into a practical measuring tool — a way of finding heights of buildings, towers, hills and the distances between far-away objects without physically reaching them. Such applications, called heights and distancesi, are used by surveyors, architects, navigators and astronomers.
Imagine a student standing near the Qutub Minar and trying to estimate its height. She cannot climb it with a measuring tape. But if she can measure how far she is from its base and the angle her gaze makes with the ground when she looks at the top, trigonometry completes the job.
Fig. 9.1 — Line of sight and angle of elevation of the top of a minar
Line of sight, Elevation and Depression
The straight line drawn from the eye of the observer to the point being viewed is called the line of sighti.
Angle of Elevation
When the object is above the horizontal level of the eye, the angle between the line of sight and the horizontal is called the angle of elevation. We raise our head to look at such an object.Fig. 9.2 — Angle of elevation α when the object is above the horizontal
Angle of Depression
When the object is below the horizontal level of the eye, the angle between the line of sight and the horizontal is called the angle of depression. We lower our head to look at such an object.Fig. 9.3 — Angle of depression β when the object is below the horizontal
In every problem of heights and distances, three steps are repeated:
- Draw a right triangle using the given geometry (the vertical object + the horizontal ground + the line of sight).
- Label the known angle (elevation or depression) and the known side.
- Choose the trigonometric ratio — sine, cosine or tangent — that connects the known quantity with the unknown.
Activity — Measure the height of your classroom pillar
Stand a known distance d (say 3 m) from the foot of a pillar. Use a protractor taped to a straw as a simple clinometer and read the angle at which you must tilt your gaze to see the top. Call it θ. Then the pillar's height above your eye-level equals \(d\,\tan\theta\). Add your eye-height to get total height.
If d = 3 m, θ = 52°, then height above eye = 3·tan 52° ≈ 3 × 1.28 ≈ 3.84 m. With eye-height 1.5 m, total ≈ 5.34 m. Your clinometer reading and tape length should be close to this result for an actual 5.3 m pillar.
Example 1 — Height of a tower
A tower stands vertically on the ground. From a point on the ground, 15 m away from the foot of the tower, the angle of elevation of the top is 60°. Find the height of the tower.
Fig. 9.4 — Right triangle ABC for the tower problem
Solution. In right triangle ABC (right-angled at B), BC = 15 m, ∠ACB = 60°, AB = h.
\[\tan 60°=\frac{AB}{BC}\Rightarrow \sqrt{3}=\frac{h}{15}\Rightarrow h=15\sqrt{3}\ \text{m}.\]So the tower is \(15\sqrt{3}\) m tall (about 25.98 m).
Example 2 — Electrician on a ladder
An electrician must repair a fault on an electric pole at a height of 5 m. He needs to reach 1.3 m below the top. The ladder, when inclined at 60° to the horizontal, should be long enough to reach the required position. How far from the foot of the pole should he place the foot of the ladder? (Take \(\sqrt{3}=1.73\).)
Fig. 9.5 — Electrician's ladder inclined at 60°
Solution. BC = 5 − 1.3 = 3.7 m (height BC where the ladder meets the pole). In right triangle BDC, ∠BDC = 60°.
\[\sin 60°=\frac{BC}{BD}\Rightarrow \frac{\sqrt3}{2}=\frac{3.7}{BD}\Rightarrow BD=\frac{2\times 3.7}{\sqrt3}\approx 4.28\ \text{m}.\] \[\cot 60°=\frac{DC}{BC}\Rightarrow \frac{1}{\sqrt3}=\frac{DC}{3.7}\Rightarrow DC=\frac{3.7}{\sqrt3}\approx 2.14\ \text{m}.\]He should place the foot of the ladder about 2.14 m from the pole; the ladder itself must be at least 4.28 m long.
Example 3 — Chimney and tall building
An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eye is 45°. Find the height of the chimney.
Fig. 9.6 — Observer's eye at AE level looking at top C
Solution. Let AB be the chimney and CD the observer. Draw DE ⊥ AB. Then DE = 28.5 m, ∠ADE = 45°, BE = CD = 1.5 m.
\[\tan 45°=\frac{AE}{DE}\Rightarrow 1=\frac{AE}{28.5}\Rightarrow AE=28.5\text{ m}.\]Height of chimney AB = AE + EB = 28.5 + 1.5 = 30 m.
Example 4 — Building and flagstaff
From a point P on the ground, the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from P. (Take \(\sqrt3=1.732\).)
Fig. 9.7 — Building AB with flagstaff BD, observer at P
Solution. Let AB = 10 m be the building, BD be the flagstaff and AP = x be the horizontal distance. In right △PAB: \(\tan 30°=AB/AP\Rightarrow 1/\sqrt3=10/x\Rightarrow x=10\sqrt3\) m ≈ 17.32 m.
In right △PAD (with AD = 10 + y, y = flagstaff length): \(\tan 45°=AD/AP\Rightarrow 1=(10+y)/10\sqrt3\Rightarrow y=10\sqrt3-10=10(\sqrt3-1)\approx 7.32\) m.
Distance ≈ 17.32 m; flagstaff ≈ 7.32 m.
Competency-Based Questions
Q1. A tree casts a shadow 20 m long when the angle of elevation of the sun is 30°. Find the height of the tree.
L3 Applytan 30° = h/20 ⇒ h = 20/√3 = 20√3/3 ≈ 11.55 m.
Q2. Riya says "The angle of elevation and the angle of depression between the tops of two equal-height buildings observed from each other's top are always different." Evaluate her claim.
L4 AnalyseWrong. For equal-height observation points, the horizontal line joins the two tops; the elevation/depression angles become zero on both sides. Even for unequal heights, the angle of elevation from the shorter building's top equals the angle of depression from the taller building's top by alternate angle theorem (horizontal lines are parallel).
Q3. A 5 m long ladder leans against a wall reaching a height of 4 m. Evaluate the angle of elevation it makes with the ground.
L5 Evaluatesin θ = 4/5 = 0.8. θ = arcsin(0.8) ≈ 53.13°. The ladder is steep but safe (not too vertical).
Q4. Design a word problem where the same pair of observations (two elevation angles) would require both tan 30° and tan 60° to be solved.
L6 CreateExample: "A tower is observed from two points on the same side. The angle of elevation from the farther point is 30° and from the nearer point is 60°. The points are 50 m apart. Find the height of the tower." Solution uses tan 60° at near point, tan 30° at far point, and subtracts horizontal distances. Answer: h = 25√3 m.
Assertion & Reason
A: The angle of elevation of an object increases when the observer moves closer to the object.
R: tan θ = height / horizontal distance, so when distance decreases, tan θ increases.
R: tan θ = height / horizontal distance, so when distance decreases, tan θ increases.
Answer: A. Since tan is increasing on (0°, 90°), a larger tan value means a larger θ.
A: From a tower's top, the angle of depression of a point on the ground equals the angle of elevation of the tower's top from that point.
R: Horizontal lines from the top and the ground point are parallel, giving equal alternate interior angles.
R: Horizontal lines from the top and the ground point are parallel, giving equal alternate interior angles.
Answer: A. The transversal (line of sight) cuts two parallels (horizontal lines at both ends), producing equal alternate angles.
A: If a 6 m pole casts a 6 m shadow, the sun's altitude is 45°.
R: When height = shadow length, tan θ = 1, so θ = 45°.
R: When height = shadow length, tan θ = 1, so θ = 45°.
Answer: A. tan 45° = 1 is the unique acute angle satisfying this.
Frequently Asked Questions
What is angle of elevation?
The angle of elevation of an object is the angle made by the line of sight with the horizontal when the object is above the observer's eye level.
What is angle of depression?
The angle of depression of an object is the angle formed by the line of sight with the horizontal when the object lies below the observer's eye level.
Are angle of elevation and depression equal for the same two points?
Yes. If A sees B at angle of elevation x, then B sees A at angle of depression x. They are alternate angles with parallel horizontals.
What trigonometric ratios are used in heights and distances?
Mainly sin, cos and tan of 30, 45, 60 and 90 degrees, applied to right triangles where the object's height is opposite the elevation angle.
Why do we use the horizontal line for these angles?
The horizontal is the natural reference for an observer standing on level ground. All elevation and depression angles are measured from it.
What must the line of sight be for right-angle trig to work?
The ground must be horizontal and the object vertical so that the line of sight forms the hypotenuse of a right triangle, allowing direct use of trig ratios.
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