Answer:
The distance between the ship at N 25°E and the lighthouse would be 7.26 miles.
Step-by-step explanation:
The question is incomplete. The complete question should be
The bearing of a lighthouse from a ship is N 37° E. The ship sails 2.5 miles further towards the south. The new bearing is N 25°E. What is the distance between the lighthouse and the ship at the new location?
Given the initial bearing of a lighthouse from the ship is N 37° E. So,
is 37°. We can see from the diagram that
would be
143°.
Also, the new bearing is N 25°E. So,
would be 25°.
Now we can find
. As the sum of the internal angle of a triangle is 180°.
![\angle ABC+\angle BCA+\angle BAC=180\\143+25+\angle BAC=180\\\angle BAC=180-143-25\\\angle BAC=12](https://tex.z-dn.net/?f=%5Cangle%20ABC%2B%5Cangle%20BCA%2B%5Cangle%20BAC%3D180%5C%5C143%2B25%2B%5Cangle%20BAC%3D180%5C%5C%5Cangle%20BAC%3D180-143-25%5C%5C%5Cangle%20BAC%3D12)
Also, it was given that ship sails 2.5 miles from N 37° E to N 25°E. We can see from the diagram that this distance would be our BC.
And let us assume the distance between the lighthouse and the ship at N 25°E is ![AC=x](https://tex.z-dn.net/?f=AC%3Dx)
We can apply the sine rule now.
![\frac{x}{sin(143)}=\frac{2.5}{sin(12)}\\ \\x=\frac{2.5}{sin(12)}\times sin(143)\\\\x=\frac{2.5}{0.207}\times 0.601\\ \\x=7.26\ miles](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7Bsin%28143%29%7D%3D%5Cfrac%7B2.5%7D%7Bsin%2812%29%7D%5C%5C%20%5C%5Cx%3D%5Cfrac%7B2.5%7D%7Bsin%2812%29%7D%5Ctimes%20sin%28143%29%5C%5C%5C%5Cx%3D%5Cfrac%7B2.5%7D%7B0.207%7D%5Ctimes%200.601%5C%5C%20%5C%5Cx%3D7.26%5C%20miles)
So, the distance between the ship at N 25°E and the lighthouse is 7.26 miles.
Answer:
question 22 ) 3.325
Step-by-step explanation:
I use the app photo math. Its free and shows you all the steps.
Answer:
2π m.
Step-by-step explanation:
Arc XY is 1/4 of the circumference of the circle
= 1/4 * 2π * r
= 1/2 π * 4
= 2π m.
Theirs your answer 661.091628959
The answer to this question is 3,3