Answer on Question #65271, Math / Geometry
You want to determine the height of the screen at a drive-in movie theater. You use a cardboard square to line up the top and bottom of the screen structure. The vertical distance from the ground to your eye is 5.5 ft and the horizontal distance from you to the screen is 14 ft. The bottom of the screen is 6 feet from the ground. Approximate the height of the screen to the nearest tenth.
Let X = X = X =
Δ A B C \Delta ABC Δ A BC
X sin θ = c sin C \frac {X}{\sin \theta} = \frac {c}{\sin C} sin θ X = sin C c
Right triangle Δ A B C \Delta ABC Δ A BC
c 2 = 1 4 2 + ( 6 − 5.5 ) 2 = 196.25 c ^ {2} = 1 4 ^ {2} + (6 - 5. 5) ^ {2} = 1 9 6. 2 5 c 2 = 1 4 2 + ( 6 − 5.5 ) 2 = 196.25
Right triangle Δ A C D \Delta ACD Δ A C D
b 2 = 1 4 2 + ( X + 6 − 5.5 ) 2 = X 2 + X + 196.25 b ^ {2} = 1 4 ^ {2} + (X + 6 - 5. 5) ^ {2} = X ^ {2} + X + 1 9 6. 2 5 b 2 = 1 4 2 + ( X + 6 − 5.5 ) 2 = X 2 + X + 196.25 sin C = 14 b \sin C = \frac {1 4}{b} sin C = b 14
Then
X = c sin θ sin C = b c sin θ 14 X = c \frac {\sin \theta}{\sin C} = b c \frac {\sin \theta}{1 4} X = c sin C sin θ = b c 14 sin θ 14 X = X 2 + X + 196.25 196.25 sin θ 1 4 X = \sqrt {X ^ {2} + X + 1 9 6 . 2 5} \sqrt {1 9 6 . 2 5} \sin \theta 14 X = X 2 + X + 196.25 196.25 sin θ 196 X 2 = 196.25 ( X 2 + X + 196.25 ) sin 2 θ 1 9 6 X ^ {2} = 1 9 6. 2 5 \left(X ^ {2} + X + 1 9 6. 2 5\right) \sin^ {2} \theta 196 X 2 = 196.25 ( X 2 + X + 196.25 ) sin 2 θ 196 196.25 sin 2 θ X 2 = X 2 + X + 196.25 \frac {1 9 6}{1 9 6 . 2 5 \sin^ {2} \theta} X ^ {2} = X ^ {2} + X + 1 9 6. 2 5 196.25 sin 2 θ 196 X 2 = X 2 + X + 196.25 0 < θ < 90 , 196 196.25 sin 2 θ > 1 ⇒ 196 > 196.25 sin 2 θ ⇒ 0 < sin θ < 14 196.25 0 < \theta < 9 0, \frac {1 9 6}{1 9 6 . 2 5 \sin^ {2} \theta} > 1 \Rightarrow 1 9 6 > 1 9 6. 2 5 \sin^ {2} \theta \Rightarrow 0 < \sin \theta < \frac {1 4}{\sqrt {1 9 6 . 2 5}} 0 < θ < 90 , 196.25 sin 2 θ 196 > 1 ⇒ 196 > 196.25 sin 2 θ ⇒ 0 < sin θ < 196.25 14 1 + cot 2 θ = 1 sin 2 θ > 196.25 196 ⇒ cot 2 θ > 196.25 196 − 1 ⇒ cot 2 θ > 0.25 196 ⇒ 1 + \cot^ {2} \theta = \frac {1}{\sin^ {2} \theta} > \frac {1 9 6 . 2 5}{1 9 6} \Rightarrow \cot^ {2} \theta > \frac {1 9 6 . 2 5}{1 9 6} - 1 \Rightarrow \cot^ {2} \theta > \frac {0 . 2 5}{1 9 6} \Rightarrow 1 + cot 2 θ = sin 2 θ 1 > 196 196.25 ⇒ cot 2 θ > 196 196.25 − 1 ⇒ cot 2 θ > 196 0.25 ⇒ = > cot θ > 1 28 = > 28 cot θ > 1 ( 196 196.25 sin 2 θ − 1 ) X 2 − X − 196.25 = 0 D = 1 − 4 ( 196 196.25 sin 2 θ − 1 ) ( − 196.25 ) = 1 − 785 + 784 sin 2 θ = 784 1 − sin 2 θ sin 2 θ = = 784 cot 2 θ X = 1 + 28 cot θ 2 ( 196 196.25 sin 2 θ − 1 ) = sin 2 θ 2 ⋅ 1 + 28 cot θ 196 196.25 − sin 2 θ ≈ sin 2 θ 2 ⋅ 1 + 28 cot θ 1 − sin 2 θ ≈ ≈ 1 2 tan 2 θ ( 1 + 28 cot θ ) f t , cot θ > 1 28 . \begin{array}{l}
=> \cot \theta > \frac{1}{28} => 28 \cot \theta > 1 \\
\left(\frac{196}{196.25 \sin^{2} \theta} - 1\right) X^{2} - X - 196.25 = 0 \\
D = 1 - 4 \left(\frac{196}{196.25 \sin^{2} \theta} - 1\right) (-196.25) = 1 - 785 + \frac{784}{\sin^{2} \theta} = 784 \frac{1 - \sin^{2} \theta}{\sin^{2} \theta} = \\
= 784 \cot^{2} \theta \\
X = \frac{1 + 28 \cot \theta}{2 \left(\frac{196}{196.25 \sin^{2} \theta} - 1\right)} = \frac{\sin^{2} \theta}{2} \cdot \frac{1 + 28 \cot \theta}{\frac{196}{196.25} - \sin^{2} \theta} \approx \frac{\sin^{2} \theta}{2} \cdot \frac{1 + 28 \cot \theta}{1 - \sin^{2} \theta} \approx \\
\approx \frac{1}{2} \tan^{2} \theta (1 + 28 \cot \theta) f t, \cot \theta > \frac{1}{28}.
\end{array} => cot θ > 28 1 => 28 cot θ > 1 ( 196.25 s i n 2 θ 196 − 1 ) X 2 − X − 196.25 = 0 D = 1 − 4 ( 196.25 s i n 2 θ 196 − 1 ) ( − 196.25 ) = 1 − 785 + s i n 2 θ 784 = 784 s i n 2 θ 1 − s i n 2 θ = = 784 cot 2 θ X = 2 ( 196.25 s i n 2 θ 196 − 1 ) 1 + 28 c o t θ = 2 s i n 2 θ ⋅ 196.25 196 − s i n 2 θ 1 + 28 c o t θ ≈ 2 s i n 2 θ ⋅ 1 − s i n 2 θ 1 + 28 c o t θ ≈ ≈ 2 1 tan 2 θ ( 1 + 28 cot θ ) f t , cot θ > 28 1 .
Answer:
the height of the screen ≈ 1 2 tan 2 θ ( 1 + 28 cot θ ) f t , cot θ > 1 28 \approx \frac{1}{2} \tan^{2} \theta (1 + 28 \cot \theta) f t, \cot \theta > \frac{1}{28} ≈ 2 1 tan 2 θ ( 1 + 28 cot θ ) f t , cot θ > 28 1
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