Chapter 4.8, Problem 26E

To determine

Tofind:the altitude of the plane using trigonometric ratio.

Answer to Problem 26E

The altitude of the plane is 3.32ft.

Explanation of Solution

Given:

Angles of elevation from $\text{C}$ to $\text{A and D}$ are $\text{1<msup><mo>6</mo><mo>′</mo></msup>}$ and $\text{5<msup><mo>4</mo><mo>′</mo></msup>}$ respectively.

Concept used:

  $\text{tanθ=}\frac{\text{perpendicular}}{\text{base}}$ it doesn’t depend on hypotenuse.

1 Hour=60 min

Calculation:

The figure can be drawn as right- angled triangle as shown below

In the above figure $\text{AB=DE=h}$ is the altitude at which the plane is flying and the angles of depression from $\text{A and D}$ are $\text{1<msup><mo>6</mo><mo>′</mo></msup>}$ and $\text{5<msup><mo>4</mo><mo>′</mo></msup>}$ respectively.

Therefore, angles of elevation from $\text{C}$ to $\text{A and D}$ are $\text{1<msup><mo>6</mo><mo>′</mo></msup>}$ and $\text{5<msup><mo>4</mo><mo>′</mo></msup>}$ respectively let $\text{BC=x}$

Use the fact that the speed of the planes is

  $\text{550miles/hr =}\frac{550}{60}\text{miles/min}$

Therefore, $\text{BE=1×}\frac{\text{55}}{\text{6}}\text{miles}$

  $\text{BE=}\frac{55}{6}\text{miles}$

In the right $\Delta \text{ABC}$ shall use the fact that $\text{θ=16}$

  $\text{tan16=}\frac{\text{AB}}{\text{BC}}$

  $\text{BC=}\frac{\text{AB}}{\text{tan16}}$ ……… (i)

  $\text{x=}\frac{\text{AB}}{\text{tan16}}$

For the right $\Delta \text{DEC}$

  $\text{tan54=}\frac{\text{DE}}{\text{EC}}$

  $\text{EC=}\frac{\text{h}}{\text{tan54}}\mathrm{..........}\text{(ii)}$

  $\text{x-}\frac{55}{6}=\frac{h}{\tan 54}$

From (i) and (ii) get

  $\frac{\text{h}}{\text{tan16}}\text{=}\frac{\text{h}}{\text{tan54}}\text{-}\frac{\text{55}}{\text{6}}$

  $\text{h=3}\text{.32}$

Therefore the altitude of the plane is 3.32ft.