91Ó°ÊÓ

In Exercise \(64,\) the Law of Cosines was used to solve a triangle in the two- solution case of SSA. Can the Law of cosines be used to solve the no-solution and single-solution cases of SSA? Explain.

The length \(s\) of a shadow cast by a vertical gnomon (a device used to tell time) of height \(h\) when the angle of the sun above the horizon is \(\theta\) can be modeled by the equation \(s=\frac{h \sin \left(90^{\circ}-\theta\right)}{\sin \theta}\) (a) Verify that the expression for \(s\) is equal to \(h \cot \theta\) (b) Use a graphing utility to complete the table. Let \(h=5\) feet. (c) Use your table from part (b) to determine the angles of the sun that result in the maximum and minimum lengths of the shadow. (d) Based on your results from part (c), what time of day do you think it is when the angle of the sun above the horizon is \(90^{\circ} ?\)

The Mach number \(M\) of a supersonic airplane is the ratio of its speed to the speed of sound. When an airplane travels faster than the speed of sound, the sound waves form a cone behind the airplane. The Mach number is related to the apex angle \(\theta\) of the cone by \(\sin (\theta / 2)=1 / M\) (a) Use a half-angle formula to rewrite the equation in terms of \(\cos \theta\). (b) Find the angle \(\theta\) that corresponds to a Mach number of 1. (c) Find the angle \(\theta\) that corresponds to a Mach number of 4.5. (d) The speed of sound is about 760 miles per hour. Determine the speed of an object with the Mach numbers from parts (b) and \((\mathrm{c})\).

The range of a projectile fired at an angle \(\theta\) with the horizontal and with an initial velocity of \(v_{0}\) feet per second is $$r=\frac{1}{32} v_{0}^{2} \sin 2 \theta$$ where \(r\) is measured in feet. An athlete throws a javelin at 75 feet per second. At what angle must the athlete throw the javelin so that the javelin travels 130 feet?

Determine whether the statement is true or false. Justify your answer. Because the sine function is an odd function, for a negative number \(u, \sin 2 u=-2 \sin u \cos u\).

The height \(h\) (in feet) above ground of a seat on a Ferris wheel at time \(t\) (in minutes) can be modeled by \(h(t)=53+50 \sin \left(\frac{\pi}{16} t-\frac{\pi}{2}\right)\) The wheel makes one revolution every 32 seconds. The ride begins when \(t=0\) (a) During the first 32 seconds of the ride, when will a person on the Ferris wheel be 53 feet above ground? (b) When will a person be at the top of the Ferris wheel for the first time during the ride? If the ride lasts 160 seconds, then how many times will a person be at the top of the ride, and at what times?