Question

From A to B a private plane flies 2.8 hours at 110 mph on a bearing of 64 degrees . It turns at point B and continues another 1.7 hours at the same speed, but on a bearing of 154 degrees to point C. Round each answer to three decimal places. a. At the end of this time, how far is the plane from its starting point? b. On what bearing is the plane from its original location?

Answer 1

Answer:

a. 360.323 m

b. $95.265^{o}$

Step-by-step explanation:

Distance covered from A to B = speed x time

                                                 = 110 x 2.8

                                                 = 308 m

Distance covered from B to C = speed x time

                                                  = 110 x 1.7

                                                  = 187 m

The sum of angles at B = $64^{o}$ + $26^{o}$

                                       = $90^{o}$

a. The distance of the plane from it starting point to C can be determined by applying the cosine rule.

$b^{2}$ = $a^{2}$ + $c^{2}$ - 2ac Cos B

Sot hat;

$b^{2}$ = $187^{2}$ + $308^{2}$ - 2(187 x 308) Cos $90^{o}$

But, Cos $90^{o}$ = 0

So that,

$b^{2}$ = $187^{2}$ + $308^{2}$

  = 34969 + 94864

  = 129833

b = $\sqrt{129833}$

  = 360.323

The distance from A to C is 360.323 m.

b. Applying the sine rule;

$\frac{a}{SinA}$ = $\frac{b}{SinB}$

$\frac{187}{SinA}$ = $\frac{360.323}{Sin90^{o} }$

$\frac{187}{SinA}$ = $\frac{360.323}{1}$

Sin A = $\frac{187}{360.323}$

         = 0.5190

⇒ A = $Sin^{-1}$ 0.5190

       = $31.265^{o}$

The bearing of the plane from its original location = $64^{o}$ + $31.265^{o}$

                                                     = $95.265^{o}$