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?

Using the Sine and Cosine rule on the triangle generated using the information given, the distance and bearing of the plane from its original location are :

Using the Cosine Rule :

b² = 187² + 308² - 2(187)(308)Cos(90°)

b² = 187² + 308² - 0

b² = 129833

b = √129833

b = 360.32 miles

B.)

Using the Sine rule :

a/sinA = b/SinB = c/SinC

187/sinA = 360.32/sin90

187/sinA = 360.32/1

sinA × 360.32 = 187

sinA = 187 / 360.32

sinA = 0.51898

A = Sin¯¹(0.51898)

A = 31.264

Hence, bearing of C from A = (64° + 31.264) = 95.26°

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a. 360.323 m

b.

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 = +

                                       =

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

= + - 2ac Cos B

Sot hat;

= + - 2(187 x 308) Cos

But, Cos = 0

So that,

= +

  = 34969 + 94864

  = 129833

b =

  = 360.323

The distance from A to C is 360.323 m.

b. Applying the sine rule;

=

=

=

Sin A =

         = 0.5190

⇒ A = 0.5190

       =

The bearing of the plane from its original location = +

                                                     =

4038

step-by-step explanation:

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