CBSE Class 10 Mathematics Standard 2026: question 38

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Step-by-step solution

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1. (i) $\angle XWA=\angle WAC=30^{\circ }$ (alternate angles, $XW\parallel AC$). In right $△AXW$, AX = d and AW = y.

   $\sin 30^{\circ }=\frac{AX}{AW}\Rightarrow \frac{1}{2}=\frac{d}{y}\Rightarrow y=2d$

   1 mark

2. (ii) $WX=AC=54$ m. In right $△BXW$:

   $\tan 45^{\circ }=\frac{BX}{WX}\Rightarrow 1=\frac{h}{54}\Rightarrow h=54$ m

   1 mark

3. (iii)(a) In right $△AXW$:

   $\tan 30^{\circ }=\frac{AX}{WX}\Rightarrow \frac{1}{\sqrt{3}}=\frac{d}{54}\Rightarrow d=\frac{54}{\sqrt{3}}=18\sqrt{3}$ m

   1 mark

4. (iii)(a) Height of the tank $AB=h+d=(54+18\sqrt{3})$ m $\approx 85.18$ m

   1 mark

5. (iii)(b) The height of the window above the ground is $WC$. Since $\angle WAC=30^{\circ }$ (alternate angles), in right $△WCA$:

   $\tan 30^{\circ }=\frac{WC}{AC}\Rightarrow \frac{1}{\sqrt{3}}=\frac{WC}{54}\Rightarrow WC=18\sqrt{3}$ m $\approx 31.18$ m

   1 mark

6. (iii)(b) In right $△BXW$:

   $\sin 45^{\circ }=\frac{h}{x}\Rightarrow \frac{1}{\sqrt{2}}=\frac{54}{x}\Rightarrow x=54\sqrt{2}$ m $\approx 76.37$ m

   1 mark

Answer: (i) y = 2d; (ii) h = 54 m; (iii)(a) (54 + 18√3) m; (iii)(b) x = 54√2 m, height of window = 18√3 m

Revise the chapter

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