Chapter 8.3, Problem 44AYU

a.

To determine

How far is it directly from Ft. Myers to Orlando.

Answer to Problem 44AYU

  $a\approx 227.38$

Explanation of Solution

Given information:

An airplane flies due north from Ft. Myers to Sarasota, a distance of 150 miles and then turns through an angle of $50°$ and flies to Orlando, a distance of 100 miles. See the figure.

  

The objective is to find how far is it directly from Ft.Myers to Orlando.

Use law of cosines formula for a triangle,

  $a^2=b^2+2bc$ Cos A,

Here a, b, c represent the sides of the triangles, and A is the angle opposite to side a.

Substitute the values 150 for b, 100 for c and $130°$ for A.

  $a^2={150}^2+{100}^2-2\left(150\right)\left(100\right)\cos 130°$

Simplify

  $a^2=22,500+10,000-30,000\left(-0.64\right)$

  $a^2$ =51700

  $a$ = $\sqrt{51700}$

  $a$

Hence, the final point is about 227.38 miles from the in initial point.

b.

To determine

What bearing should the pilot use to fly directly from Ft.Myers to orlando?

Answer to Problem 44AYU

  $\approx 19.75°$

Explanation of Solution

The objective is to find C.

Use Law of Cosines formula for a triangle, $c^2=a^2+b^2-2ab$

  $\cos C$,

  $\cos C=\frac{a^2+b^2-c^2}{2ab}$

Substitute the values and simplify.

  $\cos C=\frac{{\left(227.38\right)}^2+{150}^2-{100}^2}{2\left(227.38\right)\left(150\right)}$

= $\frac{51701.66+22,500-10,000}{68214}$

= $\frac{64201.66}{68214}$

  $\approx 0.9412$

  $C={\cos }^{-1}\left(0.9412\right)$

  $\approx 19.75°$

Hence, thepilotshoulduseabearingof $19.75°$ to fly directly from the initial point to the final point.