One evening, Kaushik was in a park. Children were playing cricket. Birds were singing on a nearby tree of height 80 m. He observed a bird on the tree at an angle of elevation of 45°. When a sixer was hit, a ball flew through the tree frightening the bird to fly away. In 2 seconds, he observed the bird flying at the same height at an angle of elevation of 30° and the ball flying towards him at the same height at an angle of elevation of 60°. (i) At what distance from the foot of the tree was he observing the bird sitting on the tree? (ii) How far did the bird fly in the mentioned time? (ii) After hitting the tree, how far did the ball travel in the sky when Kaushik saw the ball? (iii) What is the speed of the bird in m/min if it had flown 20(√(3)+1) m?

MathsCBSE Class 10 · Standard · Sample paper, session 2023-24 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard sample paper 2023-24: question 38

From the official CBSE sample question paper for 2023-24, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry

One evening, Kaushik was in a park. Children were playing cricket. Birds were singing on a nearby tree of height 80 m. He observed a bird on the tree at an angle of elevation of $4 5^{\circ}$ . When a sixer was hit, a ball flew through the tree frightening the bird to fly away. In 2 seconds, he observed the bird flying at the same height at an angle of elevation of $3 0^{\circ}$ and the ball flying towards him at the same height at an angle of elevation of $6 0^{\circ}$ .

(i)
At what distance from the foot of the tree was he observing the bird sitting on the tree?
1 mark
(ii)
How far did the bird fly in the mentioned time?
2 marks
OR
(ii)
After hitting the tree, how far did the ball travel in the sky when Kaushik saw the ball?
2 marks
(iii)
What is the speed of the bird in m/min if it had flown $20 (3 + 1)$ m?
1 mark

Figure in the paper: Diagram of Kaushik at ground point C. Tree AB is 80 m tall, with B on the ground at the foot of the tree. A bird sits at the top A. Another bird is at D, at the same height 80 m, above point E on the ground. Ball is at F, at the same height, above ground point G, with FG vertical and G directly below F; the line from C to F makes $6 0^{\circ}$ with the ground, the line CA makes $4 5^{\circ}$ and the line CD makes $3 0^{\circ}$ with the ground. Labels: Tree 80 m, Height 80 m, Ground level, points C, G, B, E on the ground line. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

Open one hint at a time and try again before the next.

Hint 1

Each object is 80 m high, so each sight line forms a right triangle with an 80 m vertical side.

Hint 2

Use tan of the angle of elevation to find each horizontal distance from Kaushik.

Hint 3

The distance moved is the difference of two horizontal distances. For speed, divide by 2 s, then multiply by 60.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) In right $△ A B C$ (tree AB = 80 m, angle at C = 45°):
$tan 4 5^{\circ} = \frac{A B}{C B} \Rightarrow 1 = \frac{80}{C B} \Rightarrow C B = 80$ m
1 mark
(ii)(a) The bird flies from A to D at the same height of 80 m, with DE = 80 m. In right $△ D E C$ :
$tan 3 0^{\circ} = \frac{D E}{C E}$
½ mark
(ii)(a) $\frac{1}{3} = \frac{80}{C E}$
½ mark
(ii)(a) $C E = 803$ m
½ mark
(ii)(a) Since $A D ∥ B E$ at the same height, the distance flown is $A D = B E = C E - C B = 803 - 80 = 80 (3 - 1)$ m $\approx 58.56$ m
½ mark
(ii)(b) The ball is at F, 80 m high above G, with the angle at C = 60°. In right $△ F GC$ :
$tan 6 0^{\circ} = \frac{F G}{C G}$
½ mark
(ii)(b) $3 = \frac{80}{C G}$
½ mark
(ii)(b) $C G = \frac{80}{3}$ m
½ mark
(ii)(b) Distance travelled by the ball after hitting the tree $= F A = GB = C B - C G = 80 - \frac{80}{3} = 80 (1 - \frac{1}{3})$ m $\approx 33.81$ m
½ mark
(iii) Speed $= \frac{distance}{time} = \frac{20 ( 3 + 1 )}{2} = 10 (3 + 1)$ m/s
½ mark
(iii) In m/min: $10 (3 + 1) \times 60 = 600 (3 + 1)$ m/min $\approx 1639.2$ m/min
½ mark

Answer: (i) 80 m; (ii)(a) 80(√3 − 1) m; (ii)(b) 80(1 − 1/√3) m; (iii) 600(√3 + 1) m/min

Where the marks are

(i) tan 45° gives CB = 80 m (1 mark)
(ii)(a) Uses tan 30° = DE/CE (½ mark)
(ii)(a) Substitutes values (½ mark)
(ii)(a) CE = 80√3 m (½ mark)
(ii)(a) Distance = CE − CB = 80(√3 − 1) m (½ mark)
(ii)(b) Uses tan 60° = FG/CG (½ mark)
(ii)(b) Substitutes values (½ mark)
(ii)(b) CG = 80/√3 m (½ mark)
(ii)(b) Distance = CB − CG = 80(1 − 1/√3) m (½ mark)
(iii) Speed in m/s = 10(√3+1) (½ mark)
(iii) Converts to 600(√3+1) m/min (½ mark)

Revise the chapter

Chapter: Some Applications of Trigonometry (notes, formulas, where marks are lost and every past question)
Topic: Line of Sight and Angle of Elevation

Practise this chapterBoard-style questions from Some Applications of Trigonometry, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2023-24 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2023-24: question 38

Hints

Step-by-step solution

Revise the chapter

More questions from Some Applications of Trigonometry