The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun's altitude is 30° than when it was 60°. Find the height of the tower and the length of original shadow. (use √(3) = 1.73)

MathsCBSE Class 10 · Basic · Board exam 2024, set 430/1/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2024: question 34(A)

Set in the CBSE Class 10 board exam of 2024, paper 430/1/1, section D, question 34(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(A) · 5 marks · Long answerSome Applications of Trigonometry

The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun's altitude is $3 0^{\circ}$ than when it was $6 0^{\circ}$ . Find the height of the tower and the length of original shadow. (use $3 = 1.73$ )

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 430/1/1.

Hints

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

Hint 1

Draw the tower with two shadows; the longer shadow corresponds to the smaller altitude $3 0^{\circ}$ .

Hint 2

Let the original shadow be x; then the longer one is x + 40. Write tan of each angle.

Hint 3

Eliminate x between the two equations to find h.

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: AB = h m is the tower (A at the foot). AC = x m is the shadow when altitude is $6 0^{\circ}$ ( $∠ B C A = 6 0^{\circ}$ ); AD = (x + 40) m is the shadow when altitude is $3 0^{\circ}$ ( $∠ B D A = 3 0^{\circ}$ ), with CD = 40 m, A, C, D collinear on the ground.
1 mark
In right $△ B A D$ : $tan 3 0^{\circ} = \frac{A B}{A D} \Rightarrow \frac{1}{3} = \frac{h}{x + 40}$
1 mark
$\Rightarrow x + 40 = h 3$ ...(i)
½ mark
In right $△ B A C$ : $tan 6 0^{\circ} = \frac{h}{x} \Rightarrow 3 = \frac{h}{x} \Rightarrow h = x 3$ ...(ii)
1 mark
Putting $x = \frac{h}{3}$ in (i): $\frac{h}{3} + 40 = h 3 \Rightarrow h (3 - \frac{1}{3}) = 40 \Rightarrow \frac{2 h}{3} = 40 \Rightarrow h = 203 = 20 \times 1.73 = 34.6$ m
1 mark
$x = \frac{h}{3} = 20$ m. Height of tower = 34.6 m, original shadow = 20 m.
½ mark

Answer: Height = $203$ = 34.6 m; original shadow = 20 m

Where the marks are

Correct labelled figure (1 mark)
$tan 3 0^{\circ} = \frac{h}{x + 40}$ (1 mark)
$x + 40 = h 3$ (½ mark)
$tan 6 0^{\circ}$ gives $h = x 3$ (1 mark)
h = $203$ = 34.6 m (1 mark)
x = 20 m (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 paper2024 board paper, set 430/1/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2024: question 34(A)

Hints

Step-by-step solution

Revise the chapter

More questions from Some Applications of Trigonometry