A spherical balloon of radius r subtends an angle of 60° at the eye of an observer. If the angle of elevation of its centre is 45° from the same point, then prove that height of the centre of the balloon is √(2) times its radius.

MathsCBSE Class 10 · Standard · Board exam 2023, set 30/1/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2023: question 33(b)

Set in the CBSE Class 10 board exam of 2023, paper 30/1/1, section D, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry

A spherical balloon of radius $r$ subtends an angle of $6 0^{\circ}$ at the eye of an observer. If the angle of elevation of its centre is $4 5^{\circ}$ from the same point, then prove that height of the centre of the balloon is $2$ times its radius.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 30/1/1.

Hints

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

Hint 1

Draw the two tangents from the eye to the balloon. The 60° is the angle between them, and the line to the centre bisects it.

Hint 2

The radius to the point of contact is perpendicular to the tangent. Use sine in that right triangle to find the distance from the eye to the centre in terms of $r$ .

Hint 3

Now use the 45° elevation in the right triangle formed by the eye, the centre and the foot of the perpendicular from the centre.

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.

Figure: Let $B$ be the observer's eye and $O$ the centre of the balloon, with radius $r$ . $B Q$ and $B P$ are the two tangents from $B$ to the balloon, touching it at $Q$ and $P$ . Join $O P$ , so $O P = r$ and $O P ⊥ B P$ . Draw $O A ⊥$ the horizontal line through $B$ . Then $∠ QB P = 6 0^{\circ}$ , $∠ A B O = 4 5^{\circ}$ , and $O A$ is the height of the centre.
1 mark
The tangents from the external point $B$ are equally inclined to $B O$ , so $B O$ bisects $∠ QB P$ :
$∠ P B O = \frac{1}{2} \times 6 0^{\circ} = 3 0^{\circ}$ .
1 mark
In right $△ B P O$ [tangent $⊥$ radius, so $∠ B P O = 9 0^{\circ}$ ]:
$sin 3 0^{\circ} = \frac{O P}{O B} = \frac{r}{O B}$ , which gives $\frac{1}{2} = \frac{r}{O B}$ , so $O B = 2 r$ ...(i)
1 mark
In right $△ B A O$ ( $∠ B A O = 9 0^{\circ}$ ):
$sin 4 5^{\circ} = \frac{O A}{O B}$ , which gives $\frac{1}{2} = \frac{O A}{O B}$ , so $O B = 2 O A$ ...(ii)
1 mark
From (i) and (ii): $2 O A = 2 r$ , so $O A = \frac{2 r}{2} = 2 r$ .
Hence the height of the centre of the balloon is $2$ times its radius. Hence proved.
1 mark

Answer: Height of centre $O A = 2 r$ (proved)

Where the marks are

Correct labelled figure with the tangents, the radius to the point of contact and the perpendicular from the centre (1 mark)
Uses the fact that BO bisects the 60° angle to get $∠ P B O = 3 0^{\circ}$ (1 mark)
Uses $sin 3 0^{\circ} = r / O B$ to get $O B = 2 r$ (1 mark)
Uses $sin 4 5^{\circ} = O A / O B$ to get $O B = 2 O A$ (1 mark)
Combines the two results to conclude $O A = 2 r$ (1 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 paper2023 board paper, set 30/1/1 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Standard 2023: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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