Question

The diagram shows a field ABCD. The bearing of B from A is 140°. C is due east of B and D is due north of C. AB = 400m, BC = 350m and CD = 450m. (a) Find the bearing of D from B.(b) Calculate the distance from D to A.

Answer 1

Step-by-step explanation:

all explanation in pics

Answer 2

Given:

The bearing of B from A is 140°.

AB= 400m

BC= 350m

CD= 450m

To Find:

The bearing from D to B

 Distance between D to A

Solution:

In ΔDBC,

                tan∅ = $\frac{P}{B}$

                         = $\frac{450}{350}$

                         = $\frac{9}{7}$

  ∅ = $tan^{-1}(\frac{9}{7}) = 52.13$

  ∅ + ∅1 = 90°

 ⇒ ∅1 = 90- ∅

          = 90- $tan^{-1}(\frac{9}{7})$

          = 90- 52.13

          = 37.87

The bearing of D from B is:

        ∅= 37.87

The distance from D to A is:

In ΔAEB,

         sin40°= $\frac{P}{H}$= $\frac{AE}{AB}$= $\frac{BE}{400}$

    ⇒ BE= 400sin°40

        cos40°= $\frac{AE}{AB}= \frac{AE}{400}$

  ⇒ AE= 400cos40°

BC= FG= 350m and AF= BE

⇒ AG= AF+ FG= 400sin40°+ 350 ..(i)

    GC= AE

⇒ GD= DC- GC= 450-400cos40° ..(ii)

                  sin40°= 0.6428

                  cos40°= 0.766

⇒ AG= 400 × 0.6428+ 350     [from(i)]

        = 607.12

⇒GD= 450- 400× 0.766    [from(ii)]

        = 143.6

 Now,

         ⇒ (AD)²= (AG)² + (GD)²

         ⇒ AD= $\sqrt{(607.12)^{2} + (143.6)^{2} }$

                   = $\sqrt{368594.69 + 20620.96}$

                   = $\sqrt{389215.65}$

                   = 623.87

                AD= 623.87