Here's how to solve part b) of the problem. Step 1: Determine the angle between the paths of Plane A and Plane B from the airport P. Plane A is on a bearing of $070^\circ$. Plane B is on a bearing of $162^\circ$. The angle $\angle APB$ is the difference between these bearings: $$ \angle APB = 162^\circ - 070^\circ = 92^\circ $$ Step 2: Calculate the actual distance between Plane A and Plane B after 3 hours. From the previous part, we know: Distance PA (from P to A) = $1200 \text{ km}$ Distance PB (from P to B) = $900 \text{ km}$ We use the Law of Cosines in $\triangle APB$ to find the distance AB: $$ AB^2 = PA^2 + PB^2 - 2(PA)(PB)\cos(\angle APB) $$ $$ AB^2 = (1200)^2 + (900)^2 - 2(1200)(900)\cos(92^\circ) $$ $$ AB^2 = 1,440,000 + 810,000 - 2,160,000\cos(92^\circ) $$ $$ AB^2 = 2,250,000 - 2,160,000(-0.034899) $$ $$ AB^2 = 2,250,000 + 75,381.84 $$ $$ AB^2 = 2,325,381.84 $$ $$ AB = \sqrt{2,325,381.84} $$ $$ AB \approx 1524.92 \text{ km} $$ Step 3: Calculate the time it takes for Plane B to reach Plane A's position. Plane B travels at its original speed, which is $300 \text{ km/h}$. $$ \text{Time} = \frac{\text{Distance}}{\text{Speed}} $$ $$ \text{Time} = \frac{1524.92 \text{ km}}{300 \text{ km/h}} $$ $$ \text{Time} \approx 5.083067 \text{ hours} $$ Step 4: Convert the time to hours and minutes, to the nearest minute. The time is $5$ full hours. To find the minutes, multiply the decimal part by $60$: $$ 0.083067 \text{ hours} \times 60 \text{ minutes/hour} \approx 4.984 \text{ minutes} $$ Rounding to the nearest minute, this is $5 \text{ minutes}$. So, the time taken is $5$ hours and $5$ minutes. Step 5: Determine the length of AB on the scale drawing. The scale is $1 \text{ cm}$ represents $200 \text{ km}$. $$ \text{Length on drawing} = \frac{\text{Actual distance}}{\text{Scale factor}} $$ $$ \text{Length on drawing} = \frac{1524.92 \text{ km}}{200 \text{ km/cm}} $$ $$ \text{Length on drawing} \approx 7.62 \text{ cm} $$ The time it takes for Plane B to reach Plane A's position is $\boxed{\text{5 hours 5 minutes}}$. The length of AB on the scale drawing is approximately $\boxed{\text{7.6 cm}}$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.