91Ó°ÊÓ

Give a general formula for all the solutions List six solutions. \(\sin \theta=\frac{\sqrt{2}}{2}\)

Find the exact value of each expression. $$ \sec \left[\sin ^{-1}\left(-\frac{1}{2}\right)\right] $$

Find the exact value, if any, of each composite function. If there is no value, state it is "not defined." Do not use a calculator. \(\sin ^{-1}\left[\sin \left(-\frac{3 \pi}{4}\right)\right]\)

Use a calculator to solve each equation on the interval \(0 \leq \theta<2 \pi .\) Round answers to two decimal places. \(\tan \theta=5\)

Solve each equation on the interval \(0 \leq \theta<2 \pi\) \((\tan \theta-1)(\sec \theta-1)=0\)

Solve the formula \(A=\frac{1}{2} b h\) for \(h\).

Use the following discussion. The formula $$ D=24\left[1-\frac{\cos ^{-1}(\tan i \tan \theta)}{\pi}\right] $$ Approximate the number of hours of daylight in New York, New York \(\left(40^{\circ} 45^{\prime}\right.\) north latitude \()\), for the following dates: (a) Summer solstice \(\left(i=23.5^{\circ}\right)\) (b) Vernal equinox \(\left(i=0^{\circ}\right)\) (c) July \(4\left(i=22^{\circ} 48^{\prime}\right)\)

Movie Theater Screens Suppose that a movie theater has a screen that is 28 feet tall. When you sit down, the bottom of the screen is 6 feet above your eye level. The angle formed by drawing a line from your eye to the bottom of the screen and another line from your eye to the top of the screen is called the viewing angle. In the figure, \(\theta\) is the viewing angle. Suppose that you sit \(x\) feet from the screen. The viewing angle \(\theta\) is given by the function $$ \theta(x)=\tan ^{-1}\left(\frac{34}{x}\right)-\tan ^{-1}\left(\frac{6}{x}\right) $$ (a) What is your viewing angle if you sit 10 feet from the screen? 15 feet? 20 feet? (b) If there are 5 feet between the screen and the first row of seats and there are 3 feet between each row and the row behind it, which row results in the largest viewing angle? (c) Using a graphing utility, graph $$ \theta(x)=\tan ^{-1}\left(\frac{34}{x}\right)-\tan ^{-1}\left(\frac{6}{x}\right) $$ What value of \(x\) results in the largest viewing angle?

Problems 83 and 84 require the following discussion: When granular materials are allowed to fall freely, they form conical (cone-shaped) piles. The naturally occurring angle, measured from the horizontal, at which the loose material comes to rest is called the angle of repose and varies for different materials. The angle of repose \(\theta\) is related to the height \(h\) and the base radius \(r\) of the conical pile by the equation \(\theta=\cot ^{-1} \frac{r}{h} .\) See the illustration. Angle of Repose: De-icing Salt Due to potential transportation issues (for example, frozen waterways), de-icing salt used by highway departments in the Midwest must be ordered early and stored for future use. When de-icing salt is stored in a pile 14 feet high, the diameter of the base of the pile is 45 feet. (a) Find the angle of repose for de-icing salt. (b) What is the base diameter of a pile that is 17 feet high? (c) What is the height of a pile that has a base diameter of approximately 122 feet?

Angle of Repose: Bunker Sand The steepness of sand bunkers on a golf course is affected by the angle of repose of the sand (a larger angle of repose allows for steeper bunkers). A freestanding pile of loose sand from a United States Golf Association (USGA) bunker had a height of 4 feet and a base diameter of approximately 6.68 feet. (a) Find the angle of repose for USGA bunker sand. Artillery A projectile fired into the first quadrant from the origin of a rectangular coordinate system will pass through the point \((x, y)\) at time \(t\) according to the relationship \(\cot \theta=\frac{2 x}{2 y+g t^{2}},\) where \(\theta=\) the angle of elevation of the launcher and \(g=\) the acceleration due to gravity \(=32.2\) feet/second \(^{2}\). An artilleryman is firing at an enemy bunker located 2450 feet up the side of a hill that is 6175 feet away. He fires a round, and exactly 2.27 seconds later he scores a direct hit. (a) What angle of elevation did he use? (b) If the angle of elevation is also given by \(\sec \theta=\frac{v_{0} t}{x}\) where \(v_{0}\) is the muzzle velocity of the weapon, find the muzzle velocity of the artillery piece he used.