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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 3 · Set 3 · Q34
The angle of elevation of the top of a building from a point A, on the ground, is $\displaystyle 30$°. On moving a distance of $\displaystyle 24$ m towards its base to the point B , the angle of elevation changes to $\displaystyle 60^{\circ}$. Find the height of the building and distance of point A from the base of the building. (Take $\displaystyle \sqrt{3}=1 \cdot 73$ )A tower stands vertically on the ground. A man standing at the top of the tower observes his friend at an angle of depression of $\displaystyle 30$°, who is approaching the foot of the tower with a uniform speed. $\displaystyle 30$ seconds later, the angle of depression changes to $\displaystyle 60$°. Find the time taken by his friend to reach the foot of the tower from this point.
The angle of elevation of the top of a building from a point A, on the ground, is $\displaystyle 30$°. On moving a distance of $\displaystyle 24$ m towards its base to the point B , the angle of elevation changes to $\displaystyle 60^{\circ}$. Find the height of the building and distance of point A from the base of the building. (Take $\displaystyle \sqrt{3}=1 \cdot 73$ )
A tower stands vertically on the ground. A man standing at the top of the tower observes his friend at an angle of depression of $\displaystyle 30$°, who is approaching the foot of the tower with a uniform speed. $\displaystyle 30$ seconds later, the angle of depression changes to $\displaystyle 60$°. Find the time taken by his friend to reach the foot of the tower from this point.
Marking-scheme solution
Let CD be the building of height 'h' m.
In right angled \(\displaystyle \Delta \mathrm{DCA}\)
\[\frac{\mathrm{h}}{x+24}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}
\]
\[\Rightarrow x+24=h \sqrt{ }
\]
In right angled \(\displaystyle \Delta \mathrm{DCB}\)
\[\frac{\mathrm{h}}{x}=\tan 60^{\circ}=\sqrt{3}
\]
\[\Rightarrow \mathrm{h}=x \sqrt{3}
\]
Solving (i) and (ii), we get
\[x=12 \text { and } \mathrm{h}=12 \sqrt{3}=12 \times 1.73=20.76
\]
\(\displaystyle \therefore \mathrm{AC}=24+12=36\)
Hence, height of the building is $\displaystyle 20.76$ m and distance of point A from the base of the building is $\displaystyle 36$ m.
Let AB be the tower of height 'h' m.
In right angled \(\displaystyle \Delta \mathrm{BAD}\)
\[\frac{\mathrm{h}}{x+y}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}
\]
In right angled \(\displaystyle \Delta \mathrm{BAC}\)
\[\frac{\mathrm{h}}{x}=\tan 60^{\circ}=\sqrt{3}
\]
Using (i) and (ii),
\[y=2 x
\]
Time taken to cover the distance \(\displaystyle y=30 \mathrm{sec}\)
∴ Time taken to cover the distance \(\displaystyle x=30 \times \frac{x}{y}=30 \times \frac{x}{2 x}=15\)
Hence, the required time taken is $\displaystyle 15$ seconds.
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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.