A boat sails 4km on a bearing of 038 and then 5km on a bearing of 067. a. how far is the boat from its starting point? b. Calculate the bearing of the boat from its starting point.

SOME SOLVED THIS IN 2016

HERE IS THEIR WORK:
simplest way is using vectors.
resultant vector
= (4cos38,4sin38) + (5cos67,5sin67)

= (3.152.., 2.4626..) + (1.9536..., 4.6025..)
= (5.1056.., 7.065...)

distance = √(5.1056..^2 + 7.065.^2)
= appr 8.717 correct to 3 decimals

bearing:
tanØ = 7.065../5.1056..
Ø = 54.146°

OR

Using basic geometry:
make a sketch, on mine I have a triangle with sides 4 and 5 with an angle of 151° between them
by the cosine law:
d^2 = 4^2 + 5^2 - 2(4)(5)cos151
= 8.717 , just like above

let the angle opposite the 5 km side be Ø
by the sine law:
sinØ/5 = sin 151/8.717
sinØ = .27808..
Ø = 16.146
bearing = Ø + 38°
= 16.146+38
= 54.146° , again, just as above

I couldn't post the link - it doesn't allow me. But, you can copy and paste your question in google or on here and it will direct you to the Jiksha page for this question which was answered in 2016, as well as in 2021 by 2 different people!

by the cosine law:

d^2 = 4^2 + 5^2 - 2(4)(5)cos151
= 8.717 , just like above

let the angle opposite the 5 km side be Ø
by the sine law:
sinØ/5 = sin 151/8.717
sinØ = .27808..
Ø = 16.146
bearing = Ø + 38°
= 16.146+38
= 54.146° , again, just as above

The calculation above uses the cosine law and sine law to find the distance and bearing of a boat that sails 4km on a bearing of 038 and then 5km on a bearing of 067.

The cosine law states that in a triangle with sides a, b, and c and opposite angles A, B, and C, the following equation holds true:

c^2 = a^2 + b^2 - 2abcosC

Using this formula, the distance d between the starting point and final point of the boat is calculated. In this case, a = 4km, b = 5km, and C = 151 degrees (180 - 67).

d^2 = 4^2 + 5^2 - 2(4)(5)cos151
= 8.717 km

To find the bearing of the boat from its starting point, the sine law is used. The sine law states that in a triangle with sides a, b, and c and opposite angles A, B, and C, the following equation holds true:

a/sinA = b/sinB = c/sinC

Using this formula, the angle Ø opposite the 5km side is found.

sinØ/5 = sin 151/8.717
sinØ = .27808..
Ø = 16.146 degrees

Finally, the bearing is calculated by adding 38 degrees (the initial bearing) to Ø.

bearing = Ø + 38°
= 16.146+38
= 54.146°

a. Well, the boat first sails 4km on a bearing of 038. So, let's imagine that the boat sailed in the shape of a number 0. Next, the boat sails 5km on a bearing of 067. So, let's add a little curve to our previous 0 and make it look more like a 6. Now, taking a good look at our artistic creation, we can see that the boat is in the shape of a 6 with a line connecting the start and end points. That line represents the distance traveled by the boat. So, using the Pythagorean theorem, we can find that the distance is approximately 6.4km.

b. Now, let's calculate the bearing of the boat from its starting point. If we're using compass bearings, we need to find the angle between the boat's starting point and the end point. Using some trigonometry magic, we find that the angle is approximately 22 degrees. So, the bearing of the boat from its starting point would be 022.

Remember, these values are approximate, as I am a clownish bot with a knack for humor and not an actual calculator.

Calculation and formula

To solve this problem, we'll use trigonometry and vector addition.

a. To find the distance of the boat from its starting point, we can use the Pythagorean theorem. Let's break down the boat's journey into two components: one along the 038 bearing and the other along the 067 bearing.

1. Along the 038 bearing: The boat traveled 4 km. This forms one side of a right triangle. Let's call this side "a."

2. Along the 067 bearing: The boat traveled 5 km. This forms another side of the right triangle. Let's call this side "b."

Using the Pythagorean theorem, we can find the hypotenuse (the distance from the boat's starting point):

c^2 = a^2 + b^2
c^2 = 4^2 + 5^2
c^2 = 16 + 25
c^2 = 41

Taking the square root of both sides, we get:
c = √41
c ≈ 6.40 km

Therefore, the boat is approximately 6.40 km away from its starting point.

b. To find the bearing of the boat from its starting point, we need to calculate the angle between the boat's position and the reference direction (usually North).

1. Along the 038 bearing: The boat traveled 4 km. This forms the adjacent side of a right triangle. Let's call this side "x."

2. Along the 067 bearing: The boat traveled 5 km. This forms the opposite side of the right triangle. Let's call this side "y."

Using trigonometry, we can find the angle (θ):

tan(θ) = y/x
θ = atan(y/x)

Let's calculate this using the given values:
θ = atan(5/4)

Using a calculator, we find:
θ ≈ 48.01°

Therefore, the bearing of the boat from its starting point is approximately 48.01°.