Question

(a) A vertical pole of height 10 m casts a shadow of 15 m on the ground and at the same time, a tower casts a shadow of 45 m on the ground. Find the height of the tower.

(b) In the given figure, $\frac{QR}{QS} = \frac{QT}{PR}$ and $∠1 = ∠2$ . Prove that $△ PQS \sim △ TQR$ .

Approach

Set up similar triangles for the pole and tower using angles of elevation. For the pole (10 m high, shadow 15 m), $tan θ = \frac{10}{15}$ . For the tower (height $h$ , shadow 45 m), $tan θ = \frac{h}{45}$ . Equating gives $h = 30$ m.

Step-by-step working

1
Draw diagrams: pole $A B = 10$ m, shadow $BC = 15$ m; tower $PQ = h$ , shadow $QR = 45$ m. Both form right triangles with the same angle of elevation.
0.5 marks
2
By similar triangles, $\frac{A B}{BC} = \frac{PQ}{QR} \Rightarrow \frac{10}{15} = \frac{h}{45}$ .
1 mark
3
Solve: $h = \frac{10 \times 45}{15} = 30$ m. Height of the tower is $30$ m.
0.5 marks

Concepts used

similar trianglesheight and shadowratioprove similarity

Approach

Step-by-step working

Concepts used

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