A ship leaves port R sails to port S and then to port T. The bearing of S from R is 112. The Bearing of T from S is 033 The distance RT is 75 km and the distance RS is 56 km
a) Draw a diagram showing the journey of the ship from R to S to T
b) calculate- i)The size of angle RST ii)The size of angle RTS iii) the bearing of R from T
c) The ship leaves port T and travels due west to a point X which is due north of R.
i) Show on your diagram the journey from T to X
ii) calculate the distance TX
PLEASE HELP ME WITH THIS ONE IT IS SUPPER IMPORTANT!!!
1. 33
2. 89
3. 31
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a) Here is a diagram showing the journey of the ship from R to S to T:
```
T
|
| 75 km
|
S
|
| 56 km
|
R
```
b) i) To find the size of angle RST, we can subtract the bearing of S from the bearing of T:
Bearing of T from S: 033°
Bearing of S from R: 112°
Angle RST = (Bearing of T from S) - (Bearing of S from R)
= 033° - 112°
= -79°
However, since angles are always measured as positive values, we can add 360° to the result:
Angle RST = -79° + 360°
= 281°
ii) To find the size of angle RTS, we subtract the size of angle RST from 180°:
Angle RTS = 180° - Angle RST
= 180° - 281°
= -101°
Again, since angles are always measured as positive values, we add 360° to the result:
Angle RTS = -101° + 360°
= 259°
iii) To find the bearing of R from T, we add 180° to the bearing of T from S:
Bearing of R from T = (Bearing of T from S) + 180°
= 033° + 180°
= 213°
c) i) Here is an updated diagram showing the ship's journey from T to X:
```
T
|
|
|
X
|
|
R
```
ii) To calculate the distance TX, we need to find the distance from T to X. Since the ship is traveling due west and X is due north of R, we know that TX is a right-angled triangle. To find the length of the hypotenuse TX, we can use the Pythagorean theorem:
TX^2 = TR^2 + RX^2
Given that the distance RT is 75 km, we need to find the distance RX. Since RX is due north of R, it will be the same as the latitude change between R and X.
Let's assume the latitude of point R is LR (in km) and the latitude of point X is LX (in km). Since TX is due west, the longitude of R and X will be the same.
Therefore, the distance RX = LX - LR.
Without the specific latitude coordinates for R and X, we cannot calculate the distance TX accurately.
a) Here is a diagram showing the journey of the ship from R to S to T:
R (start)
-->
112° / \ 033°
/ \
S T
b) i) To find the size of angle RST, we can subtract the bearing of S from R from the bearing of T from R.
Angle RST = Bearing of T from R - Bearing of S from R
= 033° - 112°
= -79° (since the clockwise direction is considered negative)
ii) To find the size of angle RTS, we can subtract the bearing of T from S from the bearing of R from S.
Angle RTS = Bearing of R from S - Bearing of T from S
= 112° - 033°
= 79° (since the clockwise direction is considered positive)
iii) To find the bearing of R from T, we need to add 180° to the bearing of T from R.
Bearing of R from T = Bearing of T from R + 180°
= 033° + 180°
= 213°
c) i) To show the journey from T to X, we need to draw a line going due west from T and then another line going due north from the point where the west line intersects the latitude of R. The latitude of R is the same as the latitude of X, as X is due north of R.
R (start)
| X
| /
| /
T ---------
ii) The distance TX can be calculated as the sum of the distance RT and the distance RX. Since the ship travels due west from T, the distance TX is the horizontal distance from T to the longitude of X.
To calculate the distance RX, we can use trigonometry. In triangle TRX, the angle RTX is a right angle since RX is due north of R. We have the length of the hypotenuse TR (75 km) and the length of one side RT (56 km). We can use the Pythagorean theorem to find the length of the other side RX.
Distance RX = √(Distance TR² - Distance RT²)
= √(75² - 56²)
= √(5625 - 3136)
= √(2489)
≈ 49.89 km
Therefore, the distance TX is approximately 56 km (distance RT) + 49.89 km (distance RX) = 105.89 km.