CBSE Class 10 Mathematics Basic 2023: question 35(b)

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1. Figure: P is the point on the ground and A is the point on the ground vertically below both planes, with PA = $x$ m. The higher plane is at C, with AC = 4000 m and $\angle CPA=60^{\circ }$. The lower plane is at B on AC, with AB = $h$ m and $\angle BPA=45^{\circ }$. Also $\angle PAC=90^{\circ }$.

   1 mark

2. In right $△PAB$: $\tan 45^{\circ }=\frac{h}{x}$, so $1=\frac{h}{x}$, i.e. $h=x$.

   1½ marks

3. In right $△PAC$: $\tan 60^{\circ }=\frac{4000}{x}$, so $x=\frac{4000}{\sqrt{3}}$ m. Hence $h=\frac{4000}{\sqrt{3}}$ m.

   1½ marks

4. Vertical distance between the planes: $BC=4000-\frac{4000}{\sqrt{3}}=4000\left(1-\frac{1}{\sqrt{3}}\right)=\frac{4000(3-\sqrt{3})}{3}$

   ½ mark

5. Using $\sqrt{3}=1.73$: $BC=\frac{4000\times 1.27}{3}=\frac{5080}{3}\approx 1693.33$ m.

   ½ mark

Answer: Vertical distance = $\frac{5080}{3}$ m ≈ 1693.33 m

Revise the chapter

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