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 and the angle of depression from the base of the tower to the same point on the opposite shore is observed to be , then the width of the river is_
(A)
(B)
© 20 m
(D)

A ball is thrown in the air in such a way that the path of the ball is modeled by the equation where represents the height of the ball in feet and the time in seconds. At what time , is the ball at its highest point?
(A) 6
(B) 2
© 3
(D) -3

If and are two points in space, what should be the value(s) of so that the distance between the two points is 3 ?
(A) and
© and
(B) and
(D) and

Ask by Phillips Nunez. in Ethiopia

Mar 09,2025

Question

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 and the angle of depression from the base of the tower to the same point on the opposite shore is observed to be , then the width of the river is_
(A)
(B)
© 20 m
(D)

A ball is thrown in the air in such a way that the path of the ball is modeled by the equation where represents the height of the ball in feet and the time in seconds. At what time , is the ball at its highest point?
(A) 6
(B) 2
© 3
(D) -3

If and are two points in space, what should be the value(s) of so that the distance between the two points is 3 ?
(A) and
© and
(B) and
(D) and

Ask by Phillips Nunez. in Ethiopia

Mar 09,2025

UpStudy AI · Accepted Answer

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 $$ a=1 $$
1. The width of the river is $$10(\sqrt{3}+1)$$ meters. 2. The ball reaches its highest point at 3 seconds. 3. The values of $$a$$ are $$\frac{5}{3}$$ and $$1$$.

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