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Mathematics · 2026 · 4 marks
CBSE 2026 · Region 5 · Set 2 · Q36
Elevated water storage tanks are built to store and supply water to nearby colonies. In the diagram given above, AB is an elevated water tank and CD is a nearby multistorey building. The building is $\displaystyle 54$ metres away from the water tank. From a window ( W ) of the building, the angle of elevation of top of the tank is $\displaystyle 45$° and angle of depression of its foot is $\displaystyle 30$°.(i)Write a relation between d (the height of window) and y. $\displaystyle 1$(ii)Determine the value of h . $\displaystyle 1$(iii)Determine height of the water tank. $\displaystyle 2$Find the value of $\displaystyle x$ and height of the window above ground level. $\displaystyle 2$
Elevated water storage tanks are built to store and supply water to nearby colonies. In the diagram given above, AB is an elevated water tank and CD is a nearby multistorey building. The building is $\displaystyle 54$ metres away from the water tank. From a window ( W ) of the building, the angle of elevation of top of the tank is $\displaystyle 45$° and angle of depression of its foot is $\displaystyle 30$°.
(i)
Write a relation between d (the height of window) and y. $\displaystyle 1$
(ii)
Determine the value of h . $\displaystyle 1$
(iii)
Determine height of the water tank. $\displaystyle 2$
Find the value of $\displaystyle x$ and height of the window above ground level. $\displaystyle 2$
Marking-scheme solution
(i)
\(\displaystyle \sin 30^{\circ}=\frac{1}{2}=\frac{\mathrm{d}}{\mathrm{y}} \Rightarrow 2 \mathrm{~d}=\mathrm{y}\)
(ii)
\(\displaystyle \tan 45^{\circ}=1=\frac{\mathrm{h}}{\mathrm{WX}}=\frac{\mathrm{h}}{54} \Rightarrow \mathrm{~h}=54 \mathrm{~m}\)
(iii)
\(\displaystyle \tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{\mathrm{d}}{54} \Rightarrow \mathrm{~d}=18 \sqrt{3} \mathrm{~m}\)
Height of the tank \(\displaystyle =\mathrm{h}+\mathrm{d}=(54+18 \sqrt{3}) \mathrm{m}\)
\(\displaystyle \angle \mathrm{WAC}=30^{\circ}, \tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{\mathrm{WC}}{54} \Rightarrow \mathrm{WC}=18 \sqrt{3} \mathrm{~m}\)
\(\displaystyle \sin 45^{\circ}=\frac{1}{\sqrt{2}}=\frac{\mathrm{h}}{\mathrm{x}} \Rightarrow \mathrm{x}=\mathrm{h} \sqrt{2} \Rightarrow \mathrm{x}=54 \sqrt{2} \mathrm{~m}\)
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CBSE Class 10 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.