324,122 views
45 votes
45 votes
Question is in the image, i need the method answer for part b.

the question says: "the diagram shows the position of two ships, A and B, and the coastguard station, C."

Question is in the image, i need the method answer for part b. the question says: &quot-example-1
User Mina Hafzalla
by
2.5k points

1 Answer

18 votes
18 votes

Explanation:

it seems like you understood the law of cosine (extended Pythagoras for non-right-angled triangles) and applied it mostly correctly for a).

but you made a little typo : suddenly, in the second half of the formula you changed the side length BC from 83.1 to 81.3. but it is 83.1.

so, the correct calculation is

a² = 83.1² + 95.5² - 2×95.5×83.1×cos(101) =

= 16,025.86 - 15,872.1×cos(101) =

= 16,025.86 - -3,028.539456... =

= 19,054.39946...

a = sqrt(19,054.39946) = 138.037674...

the rounded result is the same (138 km), but just for your information, and also, I want to use the correct numbers for b)

for b) we need to know :

bearings are angles, measured clockwise from north (north would be 0°).

when ship A looks to station C, the bearing is 146°.

when ship A looks to ship B, the corresponding bearing is 146 - the triangle angle at A (because that is how far we have to turn from looking at C to looking at B).

so, now, we need to apply the law of cosine again. this time to get angle A, as we have now side a.

c² = a² + b² - 2ab×cos(C)

and we want to get C, means now (since you called the baseline "a" in the first part)

83.1² = 95.5² + a² - 2×a×95.5×cos(angle A)

6,905.61 = 9,120.25 + 19,054.39946... - 2×138.037674...×95.5×cos(angle A)

6,905.61 = 28,174.64946... - 26,365.19574...×cos(angle A)

-21,269.03946... = -26,365.19574...×cos(angle A)

cos(angle A) = -21,269.03946.../-26,365.19574...

= 0.806708953...

angle A = 36.22437773...°

therefore, the bearing from A to B is

146 - 36.22437773... = 109.7756223...°

User Sbking
by
2.8k points