Chapter 9 — Some Applications of Trigonometry

Concept & Theory Used

• The situation forms a right-angled triangle.
• The rope acts as the hypotenuse.
• The height of the pole is the perpendicular.
• Using trigonometric ratio:

\[\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}\]

• Here, angle is given with ground → use sine ratio.

Solution Roadmap

1. Identify triangle components (hypotenuse, perpendicular).
2. Select appropriate trigonometric ratio.
3. Substitute given values.
4. Solve algebraically step-by-step.

Solution:

Length of rope (Hypotenuse) = 20 m
Angle with ground = 30°
Let height of pole = h

In right triangle,

\[\sin 30^\circ = \frac{\text{Perpendicular}}{\text{Hypotenuse}}\] \[\sin 30^\circ = \frac{h}{20}\]

Substitute value of $\sin 30^\circ = \frac{1}{2}$

\[\frac{1}{2} = \frac{h}{20}\]

Multiply both sides by 20:

\[h = 20 \times \frac{1}{2}\] \[h = 10\]

Height of the pole = 10 m

Why this Question is Important

• CBSE Board Exams: Direct application of trigonometric ratios (1–2 marks guaranteed type).
• Concept Building: Helps in identifying triangle components correctly.
• Competitive Exams (JEE/NTSE/SSC): Forms the base for height-distance problems.
• Real-Life Application: Used in measuring inaccessible heights like poles, towers, trees.

Concept & Theory Used

• The situation forms a right-angled triangle.
• The broken part of the tree acts as the hypotenuse (AC).
• The remaining vertical part is the perpendicular (BC).
• The horizontal distance is the base (AB).
• Total height of tree = AC + BC

\[\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}}, \quad \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}\]

Solution Roadmap

1. Use cosine ratio to find hypotenuse (AC).
2. Use sine ratio to find vertical part (BC).
3. Add both parts to get total height.

Solution:

Angle made by broken part with ground = 30°
AB = 8 m
Let total height of tree = h

In right triangle ABC:

Step 1: Find AC (broken part)

\[\cos 30^\circ = \frac{AB}{AC}\] \[\cos 30^\circ = \frac{8}{AC}\]

Substitute value of $\cos 30^\circ = \frac{\sqrt{3}}{2}$

\[\frac{\sqrt{3}}{2} = \frac{8}{AC}\]

Cross multiply:

\[AC = \frac{8 \times 2}{\sqrt{3}}\] \[AC = \frac{16}{\sqrt{3}} \tag{1}\]

Step 2: Find BC (vertical part)

\[\sin 30^\circ = \frac{BC}{AC}\] \[\frac{1}{2} = \frac{BC}{\frac{16}{\sqrt{3}}}\]

Multiply both sides:

\[BC = \frac{1}{2} \times \frac{16}{\sqrt{3}}\] \[BC = \frac{8}{\sqrt{3}} \tag{2}\]

Step 3: Total height of tree

\[h = AC + BC\] \[h = \frac{16}{\sqrt{3}} + \frac{8}{\sqrt{3}}\] \[h = \frac{24}{\sqrt{3}}\]

Rationalizing denominator:

\[h = \frac{24}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\] \[h = \frac{24\sqrt{3}}{3}\] \[h = 8\sqrt{3}\]

Height of tree = \(8\sqrt{3}\) m

Why this Question is Important

• CBSE Board Exams: Classic 3-step structured problem (2–4 marks).
• Concept Mastery: Combines sine and cosine in one problem.
• Competitive Exams: Frequently asked in height-distance case studies.
• Critical Skill: Teaches decomposition of complex geometry into parts.

Concept & Theory Used

• Each slide forms a right-angled triangle.
• The slide length = hypotenuse.
• The height = perpendicular.
• Use sine ratio:

\[\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}\]

Solution Roadmap

1. Apply sine formula for each slide.
2. Substitute given height and angle.
3. Solve algebraically.

Solution:

Case 1: Slide for children below 5 years

Height = 1.5 m
Angle = 30°
Let length of slide = L₁

\[\sin 30^\circ = \frac{1.5}{L_1}\] \[\frac{1}{2} = \frac{1.5}{L_1}\]

Cross multiply:

\[L_1 = 1.5 \times 2\] \[L_1 = 3 \text{ m}\]

Case 2: Slide for elder children

Height = 3 m
Angle = 60°
Let length of slide = L₂

\[\sin 60^\circ = \frac{3}{L_2}\] \[\frac{\sqrt{3}}{2} = \frac{3}{L_2}\]

Cross multiply:

\[L_2 = \frac{3 \times 2}{\sqrt{3}}\] \[L_2 = \frac{6}{\sqrt{3}}\]

Rationalizing denominator:

\[L_2 = \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\] \[L_2 = \frac{6\sqrt{3}}{3}\] \[L_2 = 2\sqrt{3} \text{ m}\]

Approximate value:

\[L_2 \approx 2 \times 1.732 = 3.464 \text{ m}\]

Lengths of slides:
Smaller slide = 3 m
Larger slide = \(2\sqrt{3}\) m ≈ 3.464 m

Why this Question is Important

• CBSE Exams: Multi-case application (very common 3–4 mark pattern).
• Concept Strength: Reinforces sine ratio usage in different scenarios.
• Competitive Exams: Helps compare geometric configurations quickly.
• Real-Life Engineering Insight: Shows relation between slope, height and length.

Concept & Theory Used

• Angle of elevation is the angle formed between the horizontal line and the line of sight when looking upwards.
• The situation forms a right-angled triangle.
• Height of tower = perpendicular, distance = base.
• Use tangent ratio:

\[\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}\]

Solution Roadmap

1. Identify base and perpendicular.
2. Apply tangent ratio.
3. Substitute values and simplify.

Solution:

Angle of elevation = 30°
Distance from foot of tower (Base) = 30 m
Let height of tower = h

Using trigonometric ratio:

\[\tan 30^\circ = \frac{\text{Perpendicular}}{\text{Base}}\] \[\tan 30^\circ = \frac{h}{30}\]

Substitute value of $\tan 30^\circ = \frac{1}{\sqrt{3}}$

\[\frac{1}{\sqrt{3}} = \frac{h}{30}\]

Multiply both sides by 30:

\[\ = \frac{30}{\sqrt{3}}\]

Rationalizing denominator:

\[\ = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\] \[\ = \frac{30\sqrt{3}}{3}\] \[\ = 10\sqrt{3}\]

Height of tower = \(10\sqrt{3}\) m

Why this Question is Important

• CBSE Board Exams: Standard height-distance problem (very high frequency).
• Core Concept: Direct use of tangent ratio.
• Competitive Exams: Foundation for multi-angle elevation problems.
• Practical Use: Used in surveying, navigation, and civil engineering.

Concept & Theory Used

• The situation forms a right-angled triangle.
• The string is the hypotenuse.
• The height of the kite is the perpendicular.
• Use sine ratio:

\[\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}\]

Solution Roadmap

1. Identify perpendicular (height) and hypotenuse (string).
2. Apply sine ratio.
3. Solve and rationalize.

Solution:

Height of kite (Perpendicular) = 60 m
Angle with ground = 60°
Let length of string = L

Using trigonometric ratio:

\[\sin 60^\circ = \frac{\text{Perpendicular}}{\text{Hypotenuse}}\] \[\sin 60^\circ = \frac{60}{L}\]

Substitute value of $\sin 60^\circ = \frac{\sqrt{3}}{2}$

\[\frac{\sqrt{3}}{2} = \frac{60}{L}\]

Cross multiply:

\[L = \frac{60 \times 2}{\sqrt{3}}\] \[L = \frac{120}{\sqrt{3}}\]

Rationalizing denominator:

\[L = \frac{120}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\] \[L = \frac{120\sqrt{3}}{3}\] \[L = 40\sqrt{3}\]

Length of the string = \(40\sqrt{3}\) m

Why this Question is Important

• CBSE Exams: Direct sine-application problem (2–3 marks).
• Concept Reinforcement: Clear understanding of hypotenuse identification.
• Competitive Exams: Base model for airborne object problems.
• Practical Insight: Used in tethered systems (kites, drones, cables).

Concept & Theory Used

• Angle of elevation is measured from eye level, not ground.
• Effective height = building height − boy’s height.
• Use tangent ratio at two positions.

\[\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}\]

Solution Roadmap

1. Find effective height.
2. Apply tan 30° to get initial distance.
3. Apply tan 60° to get final distance.
4. Subtract to find distance travelled.

Solution:

Height of building = 30 m
Height of boy = 1.5 m

Step 1: Effective height

\[\text{Effective height} = 30 - 1.5 = 28.5 \text{ m}\]

Step 2: Initial position (angle = 30°)

Let initial distance = \(d_1\)

\[\tan 30^\circ = \frac{28.5}{d_1}\] \[\frac{1}{\sqrt{3}} = \frac{28.5}{d_1}\] \[d_1 = 28.5\sqrt{3}\]

Step 3: Final position (angle = 60°)

Let final distance = \(d_2\)

\[\tan 60^\circ = \frac{28.5}{d_2}\] \[\sqrt{3} = \frac{28.5}{d_2}\] \[d_2 = \frac{28.5}{\sqrt{3}}\]

Rationalizing:

\[d_2 = \frac{28.5\sqrt{3}}{3} = 9.5\sqrt{3}\]

Step 4: Distance walked

\[\text{Distance} = d_1 - d_2\] \[= 28.5\sqrt{3} - 9.5\sqrt{3}\] \[= (28.5 - 9.5)\sqrt{3}\] \[= 19\sqrt{3}\]

Approximate value:

\[19\sqrt{3} \approx 19 \times 1.732 = 32.9 \text{ m}\]

Distance walked = \(19\sqrt{3}\) m ≈ 32.9 m

Why this Question is Important

• CBSE Exams: Classic 2-angle comparison problem (very high probability).
• Concept Depth: Introduces variable distance handling.
• Competitive Exams: Frequently appears in multi-position geometry.
• Critical Insight: Always adjust height when observer is above ground.

Concept & Theory Used

• Same observation point → same horizontal distance.
• Use tangent ratio for both angles.
• Total height = building height + tower height.

\[\tan \theta = \frac{\text{Height}}{\text{Base}}\]

Solution Roadmap

1. Use tan 45° to find horizontal distance.
2. Use tan 60° for total height.
3. Subtract building height to get tower height.

Solution:

Height of building = 20 m
Let height of tower = x m
Let horizontal distance from observation point = d

Step 1: Using angle 45° (to bottom of tower)

\[\tan 45^\circ = \frac{20}{d}\] \[1 = \frac{20}{d}\] \[d = 20 \text{ m}\]

Step 2: Using angle 60° (to top of tower)

\[\tan 60^\circ = \frac{20 + x}{d}\] \[$\sqrt{3} = \frac{20 + x}{20}\]

Multiply both sides by 20:

\[20\sqrt{3} = 20 + x\]

Rearranging:

\[x = 20\sqrt{3} - 20\] \[x = 20(\sqrt{3} - 1)\]

Height of the tower = \(20(\sqrt{3} - 1)\) m

Why this Question is Important

• CBSE Exams: Dual-angle same-point problem (very common 4-mark case).
• Concept Strength: Teaches linking two trigonometric equations.
• Competitive Exams: Frequently appears in tower + building setups.
• Key Insight: Horizontal distance remains constant.

Concept & Theory Used

• Same observation point → same horizontal distance.
• Use tangent ratio for both elevations.
• Total height = pedestal + statue.

\[\tan \theta = \frac{\text{Height}}{\text{Base}}\]

Solution Roadmap

1. Use tan 45° to express base in terms of x.
2. Use tan 60° for total height.
3. Solve equation and rationalize.

Solution:

Height of statue = 1.6 m
Let height of pedestal = x m
Let horizontal distance = d

Step 1: Using angle 45° (top of pedestal)

\[\tan 45^\circ = \frac{x}{d}\] \[1 = \frac{x}{d}\] \[d = x\]

Step 2: Using angle 60° (top of statue)

\[\tan 60^\circ = \frac{x + 1.6}{d}\] \[\sqrt{3} = \frac{x + 1.6}{x}\]

Multiply both sides by x:

\[\sqrt{3}x = x + 1.6\] \[\sqrt{3}x - x = 1.6\] \[x(\sqrt{3} - 1) = 1.6\]

Step 3: Solve for x

\[x = \frac{1.6}{\sqrt{3} - 1}\]

Rationalizing denominator:

\[x = \frac{1.6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}\] \[x = \frac{1.6(\sqrt{3} + 1)}{3 - 1}\] \[x = \frac{1.6(\sqrt{3} + 1)}{2}\] \[x = 0.8(\sqrt{3} + 1)\]

Height of pedestal = \(0.8(\sqrt{3} + 1)\) m

Why this Question is Important

• CBSE Exams: Classic two-angle, same-base application.
• Concept Strength: Reinforces equation formation using tangent.
• Competitive Exams: Appears in layered height problems.
• Critical Skill: Correct rationalization and algebra handling.

Concept & Theory Used

• Two different observation points → same horizontal distance.
• Use tangent ratio in both directions.
• Common base distance links both equations.

\[\tan \theta = \frac{\text{Height}}{\text{Base}}\]

Solution Roadmap

1. Use tan 60° from building to tower to find distance.
2. Use tan 30° from tower to building.
3. Solve for building height.

Solution:

Height of tower = 50 m
Let horizontal distance between tower and building = d
Let height of building = h

Step 1: From foot of building (angle = 60°)

\[\tan 60^\circ = \frac{50}{d}\] \[\sqrt{3} = \frac{50}{d}\] \[d = \frac{50}{\sqrt{3}}\]

Step 2: From foot of tower (angle = 30°)

\[\tan 30^\circ = \frac{h}{d}\] \[\frac{1}{\sqrt{3}} = \frac{h}{\frac{50}{\sqrt{3}}}\]

Simplify RHS:

\[\frac{1}{\sqrt{3}} = \frac{h\sqrt{3}}{50}\]

Multiply both sides by 50:

\[\frac{50}{\sqrt{3}} = h\sqrt{3}\]

Divide both sides by √3:

\[h = \frac{50}{\sqrt{3} \times \sqrt{3}}\] \[h = \frac{50}{3}\] \[h = 16\frac{2}{3} \text{ m}\]

Height of building = \( \frac{50}{3} \) m

Why this Question is Important

• CBSE Exams: Reverse-angle problem (moderate difficulty, 3–4 marks).
• Concept Depth: Requires interpreting angles from two different points.
• Competitive Exams: Tests equation linking and simplification.
• Key Insight: Same base distance is the bridge between both equations.

Concept & Theory Used

• Two right triangles share the same height.
• Different angles → different base distances.
• Use tangent ratio for both cases.

\[\tan \theta = \frac{\text{Height}}{\text{Base}}\]

Solution Roadmap

1. Assume one distance = y, other = 80 − y.
2. Apply tan 60° and tan 30°.
3. Equate both expressions of height.
4. Solve for y, then find height.

Solution:

Width of road = 80 m
Let height of each pole = x m
Let distance from point to one pole = y m
Then distance to other pole = (80 − y) m

Step 1: Using angle 60°

\[\tan 60^\circ = \frac{x}{y}\] \[\sqrt{3} = \frac{x}{y}\] \[x = y\sqrt{3} \tag{1}\]

Step 2: Using angle 30°

\[\tan 30^\circ = \frac{x}{80 - y}\] \[\frac{1}{\sqrt{3}} = \frac{x}{80 - y}\] \[x = \frac{80 - y}{\sqrt{3}} \tag{2}\]

Step 3: Equate (1) and (2)

\[y\sqrt{3} = \frac{80 - y}{\sqrt{3}}\]

Multiply both sides by √3:

\[3y = 80 - y\] \[4y = 80\] \[y = 20 \text{ m}\]

Other distance:

\[80 - y = 60 \text{ m}\]

Step 4: Find height

\[x = y\sqrt{3}\] \[x = 20\sqrt{3} \text{ m}\]

Height of poles = \(20\sqrt{3}\) m
Distances = 20 m and 60 m

Why this Question is Important

• CBSE Exams: Multi-equation trigonometry (high scoring 4-mark question).
• Concept Depth: Links two triangles via a common height.
• Competitive Exams: Tests algebra + trigonometric substitution.
• Key Insight: Larger angle → shorter distance, smaller angle → longer distance.

Concept & Theory Used

• Points lie on the same straight line from the tower.
• Second point is farther → larger base distance.
• Use tangent ratio at both positions.

\[\tan \theta = \frac{\text{Height}}{\text{Base}}\]

Solution Roadmap

1. Let canal width = x.
2. Second point distance = x + 20.
3. Apply tan 60° and tan 30°.
4. Equate heights and solve.

Solution:

Let width of canal = x m
Let height of tower = h m

Step 1: From first point (angle = 60°)

\[\tan 60^\circ = \frac{h}{x}\] \[\sqrt{3} = \frac{h}{x}\] \[h = x\sqrt{3} \tag{1}\]

Step 2: From second point (angle = 30°)

\[\tan 30^\circ = \frac{h}{x + 20}\] \[\frac{1}{\sqrt{3}} = \frac{h}{x + 20}\] \[h = \frac{x + 20}{\sqrt{3}} \tag{2}\]

Step 3: Equate (1) and (2)

\[x\sqrt{3} = \frac{x + 20}{\sqrt{3}}\]

Multiply both sides by √3:

\[3x = x + 20\] \[2x = 20\] \[x = 10 \text{ m}\]

Step 4: Find height

\[h = x\sqrt{3}\] \[h = 10\sqrt{3} \text{ m}\]

Width of canal = 10 m
Height of tower = \(10\sqrt{3}\) m

Why this Question is Important

• CBSE Exams: Classic 2-position linear problem (very frequent).
• Concept Strength: Teaches distance increment logic (x → x+20).
• Competitive Exams: Base for river/canal/tower problems.
• Key Insight: Smaller angle → larger distance.

Concept & Theory Used

• Angle of depression = angle of elevation from the ground.
• Horizontal distance remains same for both observations.
• Total height = building height + additional tower height.

\[\tan \theta = \frac{\text{Height}}{\text{Base}}\]

Solution Roadmap

1. Use angle of depression (45°) to find horizontal distance.
2. Use angle of elevation (60°) to find extra height.
3. Add building height to get total tower height.

Solution:

Height of building = 7 m
Let horizontal distance = d m
Let extra height of tower above building = y m

Step 1: Using angle of depression (45°)

\[\tan 45^\circ = \frac{7}{d}\] \[1 = \frac{7}{d}\] \[d = 7 \text{ m}\]

Step 2: Using angle of elevation (60°)

\[\tan 60^\circ = \frac{y}{d}\] \[\sqrt{3} = \frac{y}{7}\] \[y = 7\sqrt{3}\]

Step 3: Total height of tower

\[\text{Height of tower} = 7 + y\] \[= 7 + 7\sqrt{3}\] \[= 7(1 + \sqrt{3})\]

Height of tower = \(7(1 + \sqrt{3})\) m

Why this Question is Important

• CBSE Exams: Classic elevation + depression combo problem.
• Concept Depth: Tests understanding of angle equivalence.
• Competitive Exams: Very common in mixed-angle geometry.
• Key Insight: Split total height into known + unknown parts.

Concept & Theory Used

• Angle of depression = angle of elevation from ships.
• Greater angle → nearer object.
• Both ships lie on same straight line.

\[\tan \theta = \frac{\text{Height}}{\text{Base}}\]

Solution Roadmap

1. Use tan 45° to find distance of nearer ship.
2. Use tan 30° to find distance of farther ship.
3. Subtract distances to get required distance.

Solution:

Height of lighthouse = 75 m

Step 1: Distance of nearer ship (angle = 45°)

\[\tan 45^\circ = \frac{75}{d_1}\] \[1 = \frac{75}{d_1}\] \[d_1 = 75 \text{ m}\]

Step 2: Distance of farther ship (angle = 30°)

\[\tan 30^\circ = \frac{75}{d_2}\] \[\frac{1}{\sqrt{3}} = \frac{75}{d_2}\] \[d_2 = 75\sqrt{3} \text{ m}\]

Step 3: Distance between ships

\[\text{Distance} = d_2 - d_1\] \[= 75\sqrt{3} - 75\] \[= 75(\sqrt{3} - 1)\]

Distance between ships = \(75(\sqrt{3} - 1)\) m

Why this Question is Important

• CBSE Exams: Angle of depression concept (very important).
• Concept Strength: Identifying nearer vs farther object.
• Competitive Exams: Common in navigation and height-distance problems.
• Key Insight: Larger angle → smaller distance.

Concept & Theory Used

• Balloon moves horizontally → height remains constant.
• Effective height is measured from girl's eye level.
• Use tangent ratio at two positions.

\[\tan \theta = \frac{\text{Height}}{\text{Base}}\]

Solution Roadmap

1. Find effective height.
2. Find initial horizontal distance (60°).
3. Find final horizontal distance (30°).
4. Subtract to get distance travelled.

Solution:

Height of balloon = 88.2 m
Height of girl = 1.2 m

Step 1: Effective height

\[\text{Effective height} = 88.2 - 1.2 = 87 \text{ m}\]

Step 2: Initial position (angle = 60°)

Let initial horizontal distance = d₁

\[\tan 60^\circ = \frac{87}{d_1}\] \[\sqrt{3} = \frac{87}{d_1}\] \[d_1 = \frac{87}{\sqrt{3}}\] \[d_1 = 29\sqrt{3} \text{ m}\]

Step 3: Final position (angle = 30°)

Let final horizontal distance = d₂

\[\tan 30^\circ = \frac{87}{d_2}\] \[\frac{1}{\sqrt{3}} = \frac{87}{d_2}\] \[d_2 = 87\sqrt{3} \text{ m}\]

Step 4: Distance travelled

\[\text{Distance} = d_2 - d_1\] \[= 87\sqrt{3} - 29\sqrt{3}\] \[= (87 - 29)\sqrt{3}\] \[= 58\sqrt{3}\]

Distance travelled = \(58\sqrt{3}\) m

Why this Question is Important

• CBSE Exams: Moving object + angle variation problem.
• Concept Strength: Reinforces horizontal motion logic.
• Competitive Exams: Common in relative motion + trigonometry.
• Key Insight: Smaller angle → larger distance.

Concept & Theory Used

• Angle of depression = angle of elevation.
• Height of tower remains constant.
• Distances are inversely proportional to tanθ.
• Uniform speed ⇒ distance ∝ time.

\[\tan \theta = \frac{\text{Height}}{\text{Base}}\]

Solution Roadmap

1. Relate distances using tan 30° and tan 60°.
2. Find ratio of distances.
3. Use uniform speed to relate time and distance.

Solution:

Let height of tower = h
Let distance at 60° = y
Then distance at 30° = 3y

Step 1: Using trigonometric ratios

\[\tan 60^\circ = \frac{h}{y} \Rightarrow h = y\sqrt{3}\] \[\tan 30^\circ = \frac{h}{3y} = \frac{1}{\sqrt{3}}\]

This confirms distance ratio:

\[\text{Initial distance} : \text{final distance} = 3y : y\]

Step 2: Distance covered in 6 seconds

\[\text{Distance} = 3y - y = 2y\] \[\text{Time} = 6 \text{ s}\] \[\text{Speed} = \frac{2y}{6} = \frac{y}{3}\]

Step 3: Time to reach tower from second position

\[\text{Remaining distance} = y\] \[\text{Time} = \frac{y}{\frac{y}{3}}\] \[= 3 \text{ seconds}\]

Time required = 3 seconds

Why this Question is Important

• CBSE Exams: Combines trigonometry with motion.
• Concept Depth: Introduces proportional reasoning.
• Competitive Exams: Very common in speed-distance problems.
• Key Insight: tan ratio gives distance ratios directly.