CBSE Class 10 Mathematics Basic 2023: question 32(b)

Hints

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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.

1. Figure: AB = 20 m is the building, BC = h m is the tower on top of it, D is the point on the ground with AD = x m; $\angle ADB=45^{\circ }$, $\angle ADC=60^{\circ }$, $\angle A=90^{\circ }$.

   1 mark

2. In right $△ABD$: $\tan 45^{\circ }=\frac{AB}{AD}=\frac{20}{x}$

   1 mark

3. $1=\frac{20}{x}\Rightarrow x=20$ m

   ½ mark

4. In right $△ACD$: $\tan 60^{\circ }=\frac{AC}{AD}=\frac{20+h}{x}$

   1 mark

5. $\sqrt{3}x=20+h\Rightarrow 20\sqrt{3}=20+h$

   ½ mark

6. $h=20(\sqrt{3}-1)$ m

   ½ mark

7. $h=20(1.73-1)=20\times 0.73=14.6$ m. Height of the tower is 14.6 m.

   ½ mark

Answer: Height of tower = $20(\sqrt{3}-1)$ m = 14.6 m

Revise the chapter

• Chapter: Some Applications of Trigonometry (notes, formulas, where marks are lost and every past question)
• Topic: Heights and Distances: One Right Triangle