CHAPTER 12 TELANGANA CLASS X MATHEMATICS - CHAPTER 12: APPLICATIONS OF TRIGONOMETRY (EXERCISE 12.1 TO 12.2 FULL SOLUTIONS WITH SVG FIGURES)

BOARD OF SECONDARY EDUCATION, TELANGANA

CLASS 10 MATHEMATICS - CHAPTER 12: APPLICATIONS OF TRIGONOMETRY

Complete Exercise-Wise Questions & Step-by-Step Solutions with Figures (Ex 12.1 - 12.2) | Muqeem Tutorial

SUBJECT: MATHEMATICS CHAPTER 12 (FULL EXERCISES WITH FIGURES) WEIGHTAGE: 6 - 8 MARKS

EXERCISE 12.1 — Single Right-Angled Triangles & Heights and Distances

Q1. A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.

Figure 12.1: Height of Tower with 60° Angle of Elevation

Solution:
Let AB be the tower of height 'h' m and C be the observation point at distance BC = 15 m.
In right △ABC, ∠ACB = 60°.
tan 60° = Opposite / Adjacent = AB / BC
√3 = h / 15 ⇒ h = 15√3 m.
Taking √3 = 1.732: h = 15 × 1.732 = 25.98 m.
👉 Height of Tower = 15√3 m ≈ 25.98 m

Q2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 6 m. Find the height of the tree.

Figure 12.2: Broken Tree Bending to Touch Ground at 30°

Solution:
Let total height of tree = x + y, where x is vertical trunk and y is broken bent part.
Distance from foot to top C = 6 m, angle ∠C = 30°.
1. tan 30° = x / 6 ⇒ 1/√3 = x / 6 ⇒ x = 6/√3 = 2√3 m.
2. cos 30° = 6 / y ⇒ √3/2 = 6 / y ⇒ y = 12/√3 = 4√3 m.
Total Height = x + y = 2√3 + 4√3 = 6√3 m.
Taking √3 = 1.732: Total Height = 6 × 1.732 = 10.39 m.
👉 Total Height of Tree = 6√3 m ≈ 10.39 m

Q3. A contractor plans to install two slides for children in a park. For children below 5 years, she prefers a slide whose top is at height 1.5 m inclined at 30° to ground, whereas for elder children, she wants a steep slide at height 3 m inclined at 60°. What should be the length of the slide in each case?

Solution:
• Case 1 (Below 5 years): Height h₁ = 1.5 m, Angle θ₁ = 30°.
sin 30° = 1.5 / L₁ ⇒ 1/2 = 1.5 / L₁ ⇒ L₁ = 3 m.

• Case 2 (Elder children): Height h₂ = 3 m, Angle θ₂ = 60°.
sin 60° = 3 / L₂ ⇒ √3/2 = 3 / L₂ ⇒ L₂ = 6/√3 = 2√3 m ≈ 3.46 m.
👉 Slide 1 Length = 3 m  |  Slide 2 Length = 2√3 m ≈ 3.46 m

Q4. An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3 m below the top of the pole to undertake the repair work. What should be the length of the ladder that she should use which, when inclined at an angle of 60° to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? (Use √3 = 1.732)

Figure 12.3: Ladder Inclined at 60° to Reach Point 1.3 m Below Pole Top

Solution:
Total Pole Height = 5 m. Reaching Height BD = 5 - 1.3 = 3.7 m.
In right △BDC, ∠C = 60°.
1. Length of Ladder (L):
sin 60° = BD / L ⇒ √3/2 = 3.7 / L ⇒ L = 7.4 / 1.732 = 4.28 m.
2. Distance from Foot of Pole (x):
cot 60° = x / 3.7 ⇒ 1/√3 = x / 3.7 ⇒ x = 3.7 / 1.732 = 2.14 m.
👉 Ladder Length = 4.28 m  |  Distance from Foot = 2.14 m

Q5. 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 eyes is 45°. What is the height of the chimney?

Solution:
Height of observer = 1.5 m, Distance = 28.5 m.
Eye level horizontal distance = 28.5 m.
Height above eye level h = 28.5 × tan 45° = 28.5 × 1 = 28.5 m.
Total Height of Chimney = 28.5 + 1.5 = 30 m.
👉 Total Height of Chimney = 30 m

EXERCISE 12.2 — Multiple Triangles, Two Angles of Elevation/Depression & Moving Objects

Q1. A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 6