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

Hints

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Step-by-step solution

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1. Figure: AB is the building with $AB=30$ m. BP is the tower on top, with $BP=h$ m. C is the point on the ground, with $\angle ACB=30^{\circ }$ and $\angle ACP=60^{\circ }$ and $\angle CAP=90^{\circ }$.

   1 mark

2. In right $△BAC$: $\tan 30^{\circ }=\frac{AB}{AC}\Rightarrow \frac{1}{\sqrt{3}}=\frac{30}{AC}\Rightarrow AC=30\sqrt{3}$ m

   1½ marks

3. In right $△PAC$: $\tan 60^{\circ }=\frac{AP}{AC}\Rightarrow \sqrt{3}=\frac{30+h}{30\sqrt{3}}\Rightarrow 30\sqrt{3}\times \sqrt{3}=30+h$

   1½ marks

4. $90=30+h\Rightarrow h=60$
   So the height of the transmission tower is 60 m.

   1 mark

Answer: Height of transmission tower = 60 m

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