Solve the following problems: 15. As shown in the figure, the compass bearing of B from pier A is S50°E and the distance between them is 110 km. C is due east of A and due north of... Solve the following problems: 15. As shown in the figure, the compass bearing of B from pier A is S50°E and the distance between them is 110 km. C is due east of A and due north of B. If a boat sails due north from B at a constant speed of 20 km/h, will it reach C within 3 hours? Explain your answer. 16. Amy walks 350 m away from her home A in the direction N32°W to a bank B. Then she walks 285 m in the direction S58°W to a bookstore C. Find: (a) the distance between A and C, (b) the compass bearing of A from C, correct to the nearest degree. 17. In the figure, two ships B and C leave port A at the same time. Ship B sails in the direction N73°E for 64 km and ship C sails in the direction S17°E for 50 km. (a) What is the distance between B and C? (b) In what direction is B from C? Read more

Steps to Solve

1. Solve question 15: Find the distance from B to C

Since C is due east of A and due north of B, triangle ABC is a right-angled triangle with the right angle at C. Also, since the bearing of B from A is S50°E, angle BAC is $50^\circ$. We know AB = 110 km.

We need to find BC:

$BC = AB \cdot \sin(50^\circ)$

$BC = 110 \cdot \sin(50^\circ)$

$BC \approx 110 \cdot 0.766$

$BC \approx 84.26 \text{ km}$

2. Solve question 15: Calculate the time taken to reach C from B

The boat sails from B to C at 20 km/h. We have the distance $BC \approx 84.26 \text{ km}$.

Time = Distance / Speed

$T = \frac{84.26}{20}$

$T \approx 4.213 \text{ hours}$

3. Solve question 15: Determine if the boat will reach C in 3 hours

Since $4.213 > 3$, the boat will not reach C within 3 hours.

4. Solve question 16(a): Find the distance AC

Amy walks from A to B (350 m) at N32°W, and then from B to C (285 m) at S58°W. The angle between AB and BC is $32^\circ + 58^\circ = 90^\circ$. Therefore, triangle ABC is a right-angled triangle, with the right angle at B.

We can use the Pythagorean theorem to find AC:

$AC = \sqrt{AB^2 + BC^2}$

$AC = \sqrt{350^2 + 285^2}$

$AC = \sqrt{122500 + 81225}$

$AC = \sqrt{203725}$

$AC \approx 451.36 \text{ m}$

5. Solve question 16(b): Find the bearing of A from C

First, find the angle BAC:

$\tan(\angle BAC) = \frac{BC}{AB} = \frac{285}{350}$

$\angle BAC = \arctan(\frac{285}{350})$

$\angle BAC \approx 39.24^\circ$

Since Amy walked N32°W from A to B, and the angle BAC is approximately $39.24^\circ$, the bearing of A from C is $32^\circ + 39.24^\circ = 91.24^\circ$ west of south. So, the bearing of A from C is S$(91.24 - 90)^\circ$W = S1.24W. Rounding to the nearest degree, we get S1°W. We can also express the bearing as East of South. Given that $\angle BAC=39.24^\circ$, then $\angle BCA = 90^\circ - 39.24^\circ \approx 50.76^\circ$. Since Amy went S58°W to reach C and the angle BCA is about $50.76^\circ$, then angle between CA and North is $(180 - 58) - 50.76 = 122 - 50.76 = 71.24^\circ$. Then, since it's in the South-East quadrant, angle is $90-71.24^\circ$. Then A is N$71^\circ$E from C

6. Solve question 17(a): Find the distance BC

Ship B sails N73°E for 64 km, and ship C sails S17°E for 50 km. The angle between their paths is $73^\circ + 17^\circ = 90^\circ$. Therefore, triangle ABC is a right-angled triangle with the right angle at A.

Use the Pythagorean theorem to find BC:

$BC = \sqrt{AB^2 + AC^2}$

$BC = \sqrt{64^2 + 50^2}$

$BC = \sqrt{4096 + 2500}$

$BC = \sqrt{6596}$

$BC \approx 81.22 \text{ km}$

7. Solve question 17(b): Find the direction of B from C

We need to find the angle ACB.

$\tan(\angle ACB) = \frac{AB}{AC} = \frac{64}{50}$

$\angle ACB = \arctan(\frac{64}{50})$

$\angle ACB \approx 52.01^\circ$

Since ship C sailed S17°E, the direction of B from C is $90^\circ - 17^\circ - 52.01^\circ = 20.99^\circ$. So B is roughly N$21^\circ$E from C.