Question

toronto mathematics academy
(14) a pilot who is heading due north spots two forests fires. the fire that is due east is at an angle of
depression of 47°. the fire that is due west is at an angle of depression of 38°. what is the distance
between the two fires, to the nearest metre, if the altitude of the plane is 2400 m?

(15) when a pendulum swings 40° from the vertical, the bob moves 20 cm horizontally and 7.3 cm
vertically. what is the length of the pendulum, to the nearest centimetre?

image of a right triangle with a 40° angle, horizontal side 20 cm, vertical side 7.3 cm

Explanation:

Problem (14)

Step 1: Understand the angle of depression

The angle of depression is equal to the angle of elevation from the fire to the plane. Let the altitude of the plane be \( h = 2400 \) m. Let the distance from the point directly below the plane to the east fire be \( x \), and to the west fire be \( y \).

For the east fire (angle of depression \( 47^\circ \)):
\( \tan(47^\circ) = \frac{h}{x} \)
So, \( x = \frac{h}{\tan(47^\circ)} \)

For the west fire (angle of depression \( 38^\circ \)):
\( \tan(38^\circ) = \frac{h}{y} \)
So, \( y = \frac{h}{\tan(38^\circ)} \)

Step 2: Calculate \( x \) and \( y \)

First, calculate \( x \):
\( \tan(47^\circ) \approx 1.0724 \)
\( x = \frac{2400}{1.0724} \approx 2238.0 \) m

Then, calculate \( y \):
\( \tan(38^\circ) \approx 0.7813 \)
\( y = \frac{2400}{0.7813} \approx 3071.8 \) m

Step 3: Distance between the two fires

Since the two fires are east and west of the point directly below the plane, the distance between them is \( x + y \).
\( \text{Distance} = 2238.0 + 3071.8 = 5309.8 \approx 5310 \) m

Step 1: Analyze the pendulum movement

When the pendulum swings \( 40^\circ \) from the vertical, let the length of the pendulum be \( L \). The vertical displacement is \( 7.3 \) cm, so the length from the new position to the vertical line is \( L - 7.3 \) cm? Wait, no. Wait, the horizontal displacement is \( 20 \) cm, vertical displacement is \( 7.3 \) cm. Actually, the pendulum forms a right triangle where the hypotenuse is \( L \), the adjacent side (vertical) is \( L - 7.3 \)? No, wait, better to use the horizontal displacement. Wait, the horizontal displacement is \( 20 \) cm, and the angle with the vertical is \( 40^\circ \). So, the horizontal displacement is opposite to the \( 40^\circ \) angle? Wait, no. Wait, the pendulum makes a \( 40^\circ \) angle with the vertical. So, the horizontal distance (opposite side) is \( 20 \) cm, and the adjacent side (vertical) is \( L \cos(40^\circ) \), and the hypotenuse is \( L \). Wait, actually, the vertical component is \( L - 7.3 \)? Wait, no, the vertical displacement is \( 7.3 \) cm, so the length from the pivot to the bob's new position is still \( L \). Wait, let's think again.

The horizontal displacement is \( 20 \) cm, which is the opposite side to the \( 40^\circ \) angle in the right triangle formed by the pendulum, the vertical line, and the horizontal line. So, \( \sin(40^\circ) = \frac{\text{horizontal displacement}}{L} \)? Wait, no. Wait, the angle with the vertical is \( 40^\circ \), so the horizontal displacement is \( L \sin(40^\circ) \), and the vertical displacement is \( L - L \cos(40^\circ) = L(1 - \cos(40^\circ)) \). We know the horizontal displacement is \( 20 \) cm, so:

\( L \sin(40^\circ) = 20 \)

Step 2: Solve for \( L \)

\( L = \frac{20}{\sin(40^\circ)} \)

Calculate \( \sin(40^\circ) \approx 0.6428 \)

\( L = \frac{20}{0.6428} \approx 31.11 \approx 31 \) cm? Wait, but wait, the vertical displacement is \( 7.3 \) cm. Let's check with the vertical displacement.

\( L(1 - \cos(40^\circ)) = 7.3 \)

\( 1 - \cos(40^\circ) \approx 1 - 0.7660 = 0.2340 \)

\( L = \frac{7.3}{0.2340} \approx 31.19 \approx 31 \) cm. So both methods give approximately 31 cm. Wait, but let's use the horizontal displacement.

\( \sin(40^\circ) \approx 0.6428 \)

\( L = \frac{20}{0.6428} \approx 31.11 \approx 31 \) cm.

Wait, but let's verify with the vertical displacement. \( L(1 - \cos(40^\circ)) = 31.11(1 - 0.7660) = 31.11(0.2340) \approx 7.28 \approx 7.3 \) cm, which matches. So that's correct.

Answer:

The distance between the two fires is approximately \(\boxed{5310}\) metres.

Problem (15)