Question

7. A bird is flying from A towards B at an angle of 35°, a point 30 km away from A. At
Bit changes its course of flight and heads towards Con a bearing of 48° and distance
32 km away.
(i) How far is B to the North of A? (ii) How far is B to the West of A?
(iii) How far is C to the North of B? (iv) How far is C to the East of B?
(sin 55° = 0.8192, cos 55° = 0.5736, sin 42° = 0.6691, cos 42° = 0.7431)​

Answer 1

i)24.58 km

ii)17.21 km

iii) 23.78 km

Explanation:

Given,

A bird is flying  from A towards B at angle of 35°, a point 30 km away from  A

so,

∠BAC = 35°

AB = 30 km

Heads towards Con a bearing of 48° an distance 32 km away

so,

∠EBD = 48°

EB = 32 km

values,

sin 55° = 0.8192

cos 55° = 0.5736

sin 42° = 0.6691

cos 42° = 0.7431

Hence,

sin$\Theta$ = $\frac{0pp}{hyp}$

cos$\Theta$ = $\frac{adj}{hyp}$

i) Far is B to the North A .

⇒sin 55° = $\frac{AC}{30}$

   AC = 0.8192 × 30

         = 24.58 km

ii) Far is B to the West of A

⇒cos 55° = $\frac{BC}{30}$

     BC = 0.5736 × 30 = 17.21 km

iii) Far is C to the East of B  

⇒ cos 42° = $\frac{DE}{32}$

  DE = 0.7431 × 32 = 23.78 km

Answer 2

Given:-

A bird is flying  from A towards B at angle of 35°

And a point 30 km away from  A

Heads towards Con a bearing of 48° an distance 32 km away.

To find:-

(i) How far is B to the North of A? (ii) How far is B to the West of A?

(iii) How far is C to the North of B? (iv) How far is C to the East of B?

Solution:-

- ∠EBD = 48°

EB = 32 km

so,∠BAC = 35°

AB = 30 km

- And the given values are-

sin 55° = 0.8192

cos 55° = 0.5736

sin 42° = 0.6691

cos 42° = 0.7431

- Hence,

cos = (adj./ hyp.)

sin = (opp./hyp.)

- i) *Far is B to the North A.*

sin 55° = (AC/30)

AC = 0.8192 × 30

      = *24.58 km*

- ii)* Far is B to the West of A.*

cos 55° = (BC/30)

BC = 0.5736 × 30

=* 17.21 km*

- iii) *Far is C to the East of B.*

cos 42° = (DE/32)

 DE = 0.7431 × 32

= *23.78 km*