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CBSE · NCERT

Class 10 Maths – Chapter 9: Some Applications of Trigonometry

Apply trigonometry to real-world height and distance problems. Using angles of elevation and depression with tan, sin, and cos ratios to find heights of buildings, towers, poles, and distances.

Exercises: 9.1·Total Questions: 16

Exercise 9.1 16 Questions – Heights & Distances

ðŸ" Key Concepts: Angle of Elevation = angle from horizontal upward to object. Angle of Depression = angle from horizontal downward to object. Use tan θ = opposite/adjacent primarily. For right triangles: sin θ = opp/hyp, cos θ = adj/hyp.
Q 1Pole & Rope

A circus artist is climbing a 20 m long rope tightly stretched from the top of a vertical pole to the ground. The angle made by the rope with the ground level is 30°. Find the height of the pole.

sin 30° = height/20 â‡' height = 20 × sin 30° = 20 × ½ = 10 m

Q 2Broken Tree

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with it. The distance from the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Let broken part (hypotenuse) = h. Standing part = p. tan 30° = p/8 â‡' p = 8/√3. sin 30° = 8/h â‡' h = 16. Total height = p+h = 8/√3+16 = (8+16√3)/√3 = 8√3 m

Q 3Tower Height

A contractor plans to install two slides for children. For children below 5 years, slide top at 1.5m height, inclined at 30°. For older children, 3m height at 60°. Find length of each slide.

Younger: sin 30° = 1.5/l₁ â‡' l₁ = 3 m. Older: sin 60° = 3/lâ‚‚ â‡' lâ‚‚ = 3/(√3/2) = 2√3 ≈ 3.46 m

Q 4–8Tower & Building

Q5: A kite is flying at 60 m above ground. String inclined at 60°. Find string length. â†' sin 60° = 60/L â‡' L = 60/(√3/2) = 40√3 m

Q6: A 1.5 m tall boy is standing at some distance from a 30 m tall building. Angle of elevation from his eyes to top increases from 30° to 60° as he walks towards it. Find distance walked. â†' x=30√3âˆ'10√3 = 20√3 m

Q8: 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 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Let pedestal = h, distance = d. tan 45° = h/d â‡' d = h. tan 60° = (h+1.6)/h = √3 â‡' h+1.6 = h√3 â‡' h(√3âˆ'1)=1.6 â‡' h = 0.8(√3+1) m

Q 9–16Mixed Problems

Q10: Two poles of equal heights are standing opposite each other on either side of the road 80 m wide. Angles of elevation from a point between them to their tops are 60° and 30°. Find height of poles and position of point. â†' h/x=tan60°=√3, h/(80âˆ'x)=tan30°=1/√3. h=20√3. x=20. Height=20√3 m, point is 20 m from pole with 60° angle.

Q12: From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. â†' tan 45°=7/AB â‡' AB=7. tan 60°=h/7 â‡' h=7√3. Tower = 7+7√3 = 7(1+√3) m

Q14: A 1.2 m tall girl spots a balloon moving with the wind at a height of 88.2 m from ground. The angle of elevation of the balloon changes from 60° to 30° after some time. Find distance travelled by balloon. â†' tan 60°=87/x₁ â‡' x₁=87/√3. tan 30°=87/xâ‚‚ â‡' xâ‚‚=87√3. Distance = 58√3 m ≈ 100.46 m

Q16: The angles of elevation of the top of a tower from two points at distances a and b (a>b) from the base and in same straight line are complementary. Prove tower height = √(ab).

Let height=h. tan θ = h/a, tan(90°âˆ'θ) = cot θ = h/b. So h/a × h/b = tan θ × cot θ = 1 â‡' h² = ab â‡' h = √(ab) âœ"

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