Math  /  Geometry

Question7  NOT TO  SCALE \begin{array}{l} \text { NOT TO } \\ \text { SCALE } \end{array}
To avoid an island, a ship travels 40 kilometres from AA to BB and then 60 kilometres from BB to CC. The bearing of BB from AA is 080080^{\circ} and angle ABCA B C is 115115^{\circ}.

Studdy Solution

STEP 1

What is this asking? A ship goes from A to B, then B to C to avoid an island.
We know the distances and some angles, and we want to find how far it is directly from A to C. Watch out! Bearings are measured clockwise from North, so don't mix them up with regular angles!
Also, remember your cosine rule!

STEP 2

1. Find the angle BAC.
2. Use the cosine rule.

STEP 3

Alright, let's **break this down**!
We know the bearing of B from A is 8080^\circ.
This means the angle between North at A and the line AB is 8080^\circ measured clockwise.

STEP 4

We also know the angle ABC is 115115^\circ.
Now, imagine extending the line AB northward from A.
This creates an angle with the North line at B.
Since North lines are parallel, the angle between the North line at B and BA is also 8080^\circ (alternate interior angles).

STEP 5

Now, the angle between North at B and BC is the exterior angle of the triangle ABC at B.
This means it's equal to the sum of the two interior opposite angles, which are BAC and ACB.
We can write this as: Angle between North at B and BC=BAC+ACB \text{Angle between North at B and BC} = \angle BAC + \angle ACB.

STEP 6

We know the angle between North at B and BA is 8080^\circ and angle ABC is 115115^\circ, so the angle between North at B and BC is 11580=35115^\circ - 80^\circ = 35^\circ measured counter-clockwise, or 36035=325360^\circ - 35^\circ = 325^\circ measured clockwise.

STEP 7

Since bearings are measured clockwise from North, the bearing of C from B is 325325^\circ.

STEP 8

Now, we know that the angle between the North line at B and BA is 8080^\circ and the angle ABC is 115115^\circ.
So, the angle between AB and BC is 115115^\circ.
We can use this to find the angle BAC.
Since the angles in a triangle add up to 180180^\circ, we have BAC+ABC+BCA=180\angle BAC + \angle ABC + \angle BCA = 180^\circ.

STEP 9

We know ABC=115\angle ABC = 115^\circ.
Let's call BAC=x\angle BAC = x.
We can express BCA\angle BCA in terms of xx.
The angle between North at A and AB is 8080^\circ.
The angle between North at B and BC is 325325^\circ.
The angle between BA and North at A is 18080=100180^\circ - 80^\circ = 100^\circ.
The angle between BC and North at B is 360325=35360^\circ - 325^\circ = 35^\circ.
So, x+115+(10035)=180x + 115^\circ + (100^\circ - 35^\circ) = 180^\circ, which simplifies to x+115+65=180x + 115^\circ + 65^\circ = 180^\circ, and then x+180=180x + 180^\circ = 180^\circ, so x=180180=0x = 180^\circ - 180^\circ = 0^\circ.
This isn't right!

STEP 10

Let's try another approach.
We know the bearing of B from A is 8080^\circ.
We also know angle ABC is 115115^\circ.
We want to find angle BAC.
Consider the North lines at A and B.
The angle between North at A and AB is 8080^\circ.
The angle between AB and BC is 115115^\circ.
The angle between North at B and BA is 18080=100180^\circ - 80^\circ = 100^\circ.
The angle between North at B and BC is 100+115=215100^\circ + 115^\circ = 215^\circ, which means the bearing of C from B is 360215=145360^\circ - 215^\circ = 145^\circ.
The angle between North at A and AC is 180(80+(180115x))=1808065+x=35+x180^\circ - (80^\circ + (180^\circ - 115^\circ - x)) = 180^\circ - 80^\circ - 65^\circ + x = 35^\circ + x.
We have 80+(180115)+CAB=18080^\circ + (180^\circ - 115^\circ) + \angle CAB = 180^\circ, so 80+65+CAB=18080^\circ + 65^\circ + \angle CAB = 180^\circ, which means CAB=180145=35\angle CAB = 180^\circ - 145^\circ = 35^\circ.

STEP 11

The **cosine rule** states that for any triangle with sides aa, bb, and cc, and the angle CC opposite side cc, we have c2=a2+b22abcos(C)c^2 = a^2 + b^2 - 2ab \cdot \cos(C).

STEP 12

In our case, we want to find the distance AC, which we'll call cc.
We have a=60a = 60 km and b=40b = 40 km, and C=ABC=115C = \angle ABC = 115^\circ.

STEP 13

**Plugging in** the values, we get: c2=602+40226040cos(115)c^2 = 60^2 + 40^2 - 2 \cdot 60 \cdot 40 \cdot \cos(115^\circ) c2=3600+16004800cos(115)c^2 = 3600 + 1600 - 4800 \cdot \cos(115^\circ)c2=52004800(0.4226)c^2 = 5200 - 4800 \cdot (-0.4226)c25200+2028.48c^2 \approx 5200 + 2028.48c27228.48c^2 \approx 7228.48c7228.48c \approx \sqrt{7228.48}c85.02c \approx 85.02

STEP 14

The distance from A to C is approximately 85.02\boldsymbol{85.02} **km**.

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