An observer (O) is located 900 feet from a building (B). The observer notices a helicopter (H) flying at a 49° angle of elevation from his line of sight. How high is the helicopter flying over the building? You must show all work and calculations to receive full credit.

[tex]\boxed{\begin{minipage}{9.4 cm}\underline{Trigonometric ratios} \\\\$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$\\\\where:\\ \phantom{ww}$\bullet$ $\theta$ is the angle. \\ \phantom{ww}$\bullet$ $\sf O$ is the side opposite the angle. \\\phantom{ww}$\bullet$ $\sf A$ is the side adjacent the angle. \\\phantom{ww}$\bullet$ $\sf H$ is the hypotenuse (the side opposite the right angle). \\\end{minipage}}[/tex]

From inspection of the given diagram, the triangle is a right triangle. Therefore, we can use the trigonometric ratios to find the height of the helicopter.

Given:

Angle = 49°
Side opposite the angle = h
Side adjacent the angle = 900 ft

Substitute the given values into the tan ratio and solve for h:

[tex]\implies \tan 49^{\circ}=\dfrac{h}{900}[/tex]

[tex]\implies h=900\tan 49^{\circ}[/tex]

[tex]\implies h=1035.33156...[/tex]

Therefore, the height of the helicopter is 1035.3 ft (nearest tenth).