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Sample Question 13:
Step 1: Assign variables. Let's set the altitude = z, distance from Alice to airplane's ground position (point right below airplane) = y and distance from Bob to airplane's ground position = x.
Step 2: Utilize the trigonometric identity \(\tan(\theta) = \frac{opposite}{adjacent}\). From Alice's viewpoint, we have \(\tan(30^\circ) = \frac{z}{y}\). Solving for y, we have \(y = z*\sqrt{3}\).
Step 3: Similarly, from Bob's viewpoint, we have \(\tan(60^\circ) = \frac{z}{x}\). Solving for x, we get \(x = \frac{z}{\sqrt{3}}\).
Step 4: We know that the squares of the distances from Alice's and Bob's perspective should add up to the square of the actual distance between them, which yields \(x^2 + y^2 = 10^2\).
Step 5: Plug the defined x and y from step 2 and 3 into step 4's equation. Solving the equation, we find \(z=\sqrt{30}\), which is approximately 5.5 miles.
Hence, the altitude of the airplane is closest to 5.5 miles, or \(\boxed{\textbf{(E)}\ 5.5}\).
Alice and Bob live \(10\) miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is \(30^\circ\) from Alice's position and \(60^\circ\) from Bob's position. Which of the following is closest to the airplane's altitude, in miles?
\(\textbf{(A)}\ 3.5 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 4.5 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 5.5\)
Answer Keys
Question 13: ESolutions
Question 13Step 1: Assign variables. Let's set the altitude = z, distance from Alice to airplane's ground position (point right below airplane) = y and distance from Bob to airplane's ground position = x.
Step 2: Utilize the trigonometric identity \(\tan(\theta) = \frac{opposite}{adjacent}\). From Alice's viewpoint, we have \(\tan(30^\circ) = \frac{z}{y}\). Solving for y, we have \(y = z*\sqrt{3}\).
Step 3: Similarly, from Bob's viewpoint, we have \(\tan(60^\circ) = \frac{z}{x}\). Solving for x, we get \(x = \frac{z}{\sqrt{3}}\).
Step 4: We know that the squares of the distances from Alice's and Bob's perspective should add up to the square of the actual distance between them, which yields \(x^2 + y^2 = 10^2\).
Step 5: Plug the defined x and y from step 2 and 3 into step 4's equation. Solving the equation, we find \(z=\sqrt{30}\), which is approximately 5.5 miles.
Hence, the altitude of the airplane is closest to 5.5 miles, or \(\boxed{\textbf{(E)}\ 5.5}\).