CBSE Class 10 Mathematics Standard 2023: question 32(B)

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Step-by-step solution

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1. Figure: Let DE be the building, with D its foot and E its top, so $DE=7$ m. Let AC be the cable tower, with C its foot and A its top, and $AC=h$ m.

   Draw EB ⟂ AC, so EB is horizontal and BCDE is a rectangle. Then $BC=DE=7$ m and $AB=(h-7)$ m.

   The angle of elevation is $\angle AEB=60^{\circ }$ and the angle of depression is $\angle BEC=30^{\circ }$. Since EB ∥ CD (alternate angles), $\angle ECD=30^{\circ }$.

   1 mark

2. Let $CD=x$ m. In right $△EDC$:

   $\tan 30^{\circ }=\frac{DE}{CD}$, so $\frac{1}{\sqrt{3}}=\frac{7}{x}$

   $x=7\sqrt{3}$ m

   So $BE=CD=7\sqrt{3}$ m.

   1½ marks

3. In right $△ABE$:

   $\tan 60^{\circ }=\frac{AB}{BE}$

   1 mark

4. $\frac{h-7}{7\sqrt{3}}=\sqrt{3}$, so $h-7=7\times 3=21$

   1 mark

5. $h=28$

   The height of the tower is 28 m.

   ½ mark

Answer: 28 m

Revise the chapter

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