A tower 20 m high is on the bank of a river. It is observed that the angle of depression from the top of the tower to the point on the opposite shore is \( 60^{\circ} \) and the angle of depression from the base of the tower to the same point on the opposite shore is observed to be \( 45^{\circ} \), then the width of the river is_ \( \qquad \) (A) \( 10 \sqrt{3} \mathrm{~m} \) (B) \( 10(\sqrt{3}+1) \mathrm{m} \) (C) 20 m (D) \( \frac{20 \sqrt{3}}{3} \mathrm{~m} \) A ball is thrown in the air in such a way that the path of the ball is modeled by the equation \( y=-x^{2}+6 x \) where \( y \) represents the height of the ball in feet and \( x \) the time in seconds. At what time \( x \), is the ball at its highest point? (A) 6 (B) 2 (C) 3 (D) -3 If \( P=(3, a-1, a+2) \) and \( Q=(2 a+1,3,3 a) \) are two points in space, what should be the value(s) of \( a \) so that the distance between the two points is 3 ? (A) \( a=-\frac{5}{3} \) and \( a=1 \) (C) \( a=\frac{5}{3} \) and \( a=1 \) (B) \( a=\frac{5}{3} \) and \( a=-1 \) (D) \( a=-\frac{5}{3} \) and \( a=-1 \)

1. Let the tower’s base be at point \( A=(0,0) \) and its top at \( T=(0,20) \). Assume the point \( P \) on the opposite shore has coordinates \( (d, y) \). We are given: - The angle of depression from \( T \) to \( P \) is \( 60^{\circ} \). - The angle of depression from \( A \) to \( P \) is \( 45^{\circ} \). Since the angle of depression is measured from the horizontal, we have: From \( T \): The line of sight makes an angle \( 60^{\circ} \) below the horizontal so that \[ \tan 60^{\circ} = \frac{20 - y}{d}. \] Thus, \[ 20 - y = d \sqrt{3}. \] From \( A \): The angle of depression is \( 45^{\circ} \). Since the horizontal through \( A \) is at \( y=0 \), \( P \) must lie \( y \) units below \( A \). Thus, \[ \tan 45^{\circ} = \frac{0 - y}{d} = \frac{-y}{d}. \] Since \(\tan 45^{\circ}=1\), it follows that \[ -y= d \quad\Longrightarrow\quad y = -d. \] Substitute \( y=-d \) in the first equation: \[ 20 - (-d) = 20 + d = d\sqrt{3}. \] Solve for \( d \): \[ 20 + d = d\sqrt{3} \quad\Longrightarrow\quad 20 = d(\sqrt{3}-1) \quad\Longrightarrow\quad d=\frac{20}{\sqrt{3}-1}. \] Rationalize the denominator: \[ d=\frac{20(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{20(\sqrt{3}+1)}{3-1}=\frac{20(\sqrt{3}+1)}{2}=10(\sqrt{3}+1). \] Thus, the width of the river is \(10(\sqrt{3}+1) \) m. 2. The ball’s height is given by \[ y=-x^2+6x. \] For a quadratic function \( y=ax^2+bx+c \) (with \( a<0 \) for a downward opening parabola), the maximum occurs at \[ x=-\frac{b}{2a}. \] Here, \( a=-1 \) and \( b=6 \): \[ x=-\frac{6}{2(-1)}=\frac{6}{2}=3. \] Hence, the ball is at its highest point at \( x=3 \) seconds. 3. The points are \[ P=(3,\, a-1,\, a+2) \quad \text{and} \quad Q=(2a+1,\, 3,\, 3a). \] The distance \( d \) between them is \[ d=\sqrt{(x_Q-x_P)^2+(y_Q-y_P)^2+(z_Q-z_P)^2}. \] Compute each coordinate difference: - \( x \)-coordinate: \[ 2a+1-3=2a-2=2(a-1). \] - \( y \)-coordinate: \[ 3-(a-1)=4-a. \] - \( z \)-coordinate: \[ 3a-(a+2)=2a-2=2(a-1). \] Then \[ d=\sqrt{[2(a-1)]^2+ (4-a)^2+ [2(a-1)]^2}. \] Simplify: \[ d=\sqrt{4(a-1)^2+(4-a)^2+4(a-1)^2}=\sqrt{8(a-1)^2+(4-a)^2}. \] We are given \( d=3 \), so: \[ 8(a-1)^2+(4-a)^2 = 9. \] Expand the terms: \[ 8(a^2-2a+1) + (a^2-8a+16) = 8a^2-16a+8+a^2-8a+16. \] Combine like terms: \[ 9a^2-24a+24 = 9. \] Rearranging: \[ 9a^2-24a+15=0. \] Divide by 3: \[ 3a^2-8a+5=0. \] Solve the quadratic using the quadratic formula: \[ a=\frac{8 \pm \sqrt{(-8)^2-4(3)(5)}}{2(3)}=\frac{8\pm \sqrt{64-60}}{6}=\frac{8\pm 2}{6}. \] This gives: \[ a=\frac{10}{6}=\frac{5}{3} \quad \text{or} \quad a=\frac{6}{6}=1. \] Therefore, \( a=\frac{5}{3} \) and \( a=1 \). Final Answers: 1. \(10(\sqrt{3}+1) \) m 2. \( x=3 \) seconds 3. \( a=\frac{5}{3} \) and \( a=1 \)