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Question
question 4 (essay worth 10 points)
(05.04 mc)
an observer (o) is located 300 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.
Step1: Use the tangent function
In a right - triangle, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Here, \(\theta = 49^{\circ}\), the adjacent side to the angle of elevation is \(300\) feet (distance from the observer to the building, \(BH\)), and the opposite side is the height of the helicopter above the building (\(h\)). So, \(\tan(49^{\circ})=\frac{h}{300}\).
Step2: Solve for \(h\)
We know that \(\tan(49^{\circ})\approx1.1504\). Then \(h = 300\times\tan(49^{\circ})\). Substitute the value of \(\tan(49^{\circ})\): \(h=300\times1.1504 = 345.12\) feet.
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The helicopter is flying at a height of \(345.12\) feet over the building.