Chapter 4.8, Problem 35E

To determine

Tofind:How far east will the plane have travelled from its point of departure.

Answer to Problem 35E

The plane is flying 709 miles east after 1.5 hours.

Explanation of Solution

Given:

the bearing of plane is ${\text{N52}}^0\text{E}$

Plane is at speed of 600 MPH at time 1.5 hours.

Concept used:

Speed $\times$ time=distance

Trigonometric ratio:

  $\text{sinθ=}\frac{\text{opposite}}{\text{hypotenuse}}$

Calculation:

Since, the bearing of plane is ${\text{N52}}^0\text{E}$ then if we consider direction east as our base, then the angle of elevation of the plane from the airport will be

  ${90}^0-{52}^0={38}^0$

Let the vertical height travelled by plane from base in time 1.5 hours be h.

Distance travelled by plane at speed of 600 MPH at time 1.5 hours is given by

Speed $\times$ time=distance

  $\Rightarrow 600\times 1.5$

  $\text{900miles}$

Now by considering right $\Delta \text{ONE}$ and use the fact that:

  ${\text{sin38}}^{\text{0}}\text{=}\frac{\text{opposite}}{\text{hypotenuse}}$

  $\Rightarrow \frac{h}{900}$

  ${\text{900sin38}}^{\text{0}}\text{=h}$

  $\text{900}\times \text{0}\text{.61566=h}$

  $\text{h=554}\text{.09miles}$

Therefore, the plane is flying 554miles north after 1.5 hours.

Let the horizontal distance travelled by the plane in 1.5 hours be l.

Again, in right $\Delta \text{ONE}$ use the fact that

  ${\text{cos38}}^{\text{0}}\text{=}\frac{\text{adjacent}}{\text{hypotenuse}}$

  $\Rightarrow \frac{l}{900}$

  $900\times \cos {38}^0=l$

  $900\times 0.7880=l$

  $\text{l=709miles}$

Therefore the plane is flying 709 miles east after 1.5 hours.