UPSC CDS 2 2021 Math Paper

Given, altitude of aeroplane is 1200 m.
and the width of the river is 3000m.
So, we have,
BC=3000m,AD=1200m,
∠B=30∘ and ∠C=θ,In △ABD,
‌tan‌30∘=‌

AD

BD

⇒‌

1

√3

=‌

1200

BD

⇒BD=1200√3
∴DC=BC−BD=3000−1200√3
ln∆ACD‌,
cot‌θ‌‌=‌

DC

AD

‌‌=‌

3000−1200√3

1200

‌‌=‌

30−12√3

12

=‌

5−2√3

2

cot‌θ‌‌=2.5−1.732=0.77
We know, cot‌θ is a decreasing function in first quadrant.
cot‌45∘=1
and cot‌60∘=‌

1

√3

=‌

√3

3

=‌

1.732

3

≈0.6
∴cot‌45∘>cot‌θ>cot‌60∘
∴θ∈(45∘,60∘)
Hence, option (c) is correct.