Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 274: 66


The takeoff angle $\theta$ is $14^\circ$.

Work Step by Step

$$0.342D\cos\theta+h\cos^2\theta=\frac{16D^2}{V_0^2}$$ The question asks for $\theta$, which is the takeoff angle, the kind of angle of takeoff which you might encounter in action movie when a car starts to fly out of control after being hit. That means the interval concerned here is $[0^\circ, 180^\circ)$. The car cannot fly at an angle smaller or larger than that. 1) Put in $V_0=60, D=80, h=2$ to the equation $$0.342\times80\cos\theta+2\cos^2\theta=\frac{16\times80^2}{60^2}$$ $$27.36\cos\theta+2\cos^2\theta=\frac{102400}{3600}$$ $$27.36\cos\theta+2\cos^2\theta\approx28.44$$ 2) Solve the equation for $\theta$: $$27.36\cos\theta+2\cos^2\theta=28.44$$ $$2\cos^2\theta+27.36\cos\theta-28.44=0$$ $$\cos^2\theta+13.68\cos\theta-14.22=0$$ We take the equation as a quadratic formula, whose $a=1, b=13.68, c=-14.22$ - Find $\Delta$: $\Delta=b^2-4ac=13.68^2-4\times1\times(-14.22)=187.1424+56.88=244.0224$ - Find $\cos\theta$: $$\cos\theta=\frac{-b\pm\sqrt\Delta}{2a}=\frac{-13.68\pm\sqrt{244.0224}}{2}$$ - For $\cos\theta=\frac{-13.68+\sqrt{244.0224}}{2}\approx0.971$ We would have $$\theta=\cos^{-1}0.971\approx14^\circ\in[0^\circ,180^\circ)$$ This solution is accepted. - For $\cos\theta=\frac{-13.68-\sqrt{244.0224}}{2}\approx-14.651$ As the range of a cosine function is $[-1,1]$, and $-14.651$ lies out of that range, there is no value of $\theta$ which $\cos\theta$ can equal $-14.651$ So overall, only one possibility of $\theta$ has been found. The takeoff angle $\theta$ is $14^\circ$.
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