This question should be answered using trigonometry, not scale drawings.

Question

An aeroplane leaves Airport A and flies due south for 352 km to
Airport B. It then flies to Airport C, which is 63 degrees east of north
of B, and 42 degrees east of south of A. Finally, it flies due south to
Airport D, which is situated due east of Airport B. (You may assume that
all distances are flat and in a straight line.)

(a) Draw a diagram showing the positions of Airports A, B,Cand D,
marking in the angles and the lengths that you are given.

(b) Find the distance between Airport B and Airport C, that is, the
length of side BC.

(c) Find the distance between Airport B and Airport D. 

(d) A passenger at Airport A wants to fly to Airport D. He can fly via
Airport C or via Airport B. If the distance from Airport A to
Airport C is 325 km (to three significant figures), which route is
shorter? Explain your answer.

Answer 1

If you draw the diagram----

<BAC=42 deg

<ABC=63 deg

Since ABC forms the triangle

 

So, <BCA=75 dig

AB =352 KM

Applying the sin formula

 

sin(75)/352  =sin(42)/BC  =sin(63)/AC

 

BC= 352*sin(42)/sin(75)

 

     =243.8 km

AC= 352*sin(63)/sin(75)

 

     =324.7

Since < ABD=90 deg given

<CBD =90-63=27 dig

 

BD =BC cos(27)

     =217.2 km

CD =BC sin(27)

     =110.7 km

 

AC+CD =435.4 km

AB+BD =569.2 km

 

So,AC+ CD is shorter