UPSC NDA Math Model Paper 8

Let's first draw the diagram of the given information: Given: AB = 1 km, ∠ MAN = 30°
, ∠ MBN = 75°
In Δ MAN, tan 30° = MN/AM
⇒ MN = AM tan 30°...(1)
And in Δ MBN, tan 75° = MN/BM
⇒ MN = BM tan 75° ......(2)
Using (1) and (2), AM tan 30° = BM tan 75°
⇒(AB+BM)

1

√3

=BM(2+√3)
⇒(1+BM)=2√3BM+3BM
⇒BM=

1

2+2√3

And, MN = BM tan75°
⇒MN=

1

2+2√3

(2+√3) =

2+√3

2+2√3

×

2−2√3

2−2√3

⇒MN=

4−4√3+2√3−6

4−12

=

2+2√3

8

⇒MN=

1

4

(√3+1)km =250(√3+1)m