A boat leaves the harbour and travels 9 km west, then 12 km north. Calculate the distance from the harbour. *
Answers
Answer:
= {9^2} + {12^2} - (2 \times 9 \times 12 \times \cos 135^\circ )
\frac{{19.4}}{{\sin 135^\circ }} = \frac{9}{{\sin x^\circ }}
180⁰ - 19.1⁰ = 159.9⁰ and.
Step-by-step explanation:
a) its distance from the harbour
b) the bearing it now has to sail to get back to the harbour
We need to sketch the problem first.
Find the size of angle R.
A path of a ship sailing 9km NW and 12km N
The boat travels from the harbour in a NW direction. This is halfway between west and north so it is at an angle of 45° to the axis for east-west and the axis for north-south. Angles of 45° are formed as shown in the diagram below.
Harbour image breaking down the main angle to 90 degrees and 45 degrees.
When the boat's direction changes to N, the internal angle formed can be found by adding 45° and 90°.
The 90° comes from the angle between the north-south and east-west axes
This angle is 135°, as shown below.
Diagram of bearings triangle with 135° angle and values 12km and 9km
The distance of the boat from the harbour is the missing side.
We know two sides and included angle, so we use the cosine rule.
Diagram of bearings triangle with angle to find
We now have to find the new bearing to sail back to the harbour.
Let' look again the sketch.,
If we find the angle then we can find our required angle by subtracting this answer from 180°. The angles in a straight line add up to 180°, in this case the north line.
We can use the sine rule or the cosine rule as we now have so much information. The sine rule is usually quicker.
Answer:
15
Step-by-step explanation:
hypotenuse 2 =Height 2 + base 2
X =9x9 +12x12
X =144 + 225
X=225 2
X =15