Question

Ship A and Ship B are 120 km apart when they pick up a distress call from another boat. Ship B estimates that they are 70 km away from the distress call. They also notice that the angle between the line from ship B to ship A and the line from ship A to the distress call is 28°. What are the two possible distances, to the nearest TENTH of a km, from ship A to the boat?

Answer 1

The two possible distance are 147.5km and 64.4km

Explanation:

Given

See Attachment for Illustration

Required

Determine the possible distance of ship A from the boat

The distress is represented by X;

So, the question requires we calculate distance AX;

This will be done using cosine formula as follows;

`a^2 = b^2 + x^2 - 2bxCosA`

In this case;

a = 70;

b = ??

x = 120

A = 28 degrees

Substitute these values in the formula above

`70^2 = b^2 + 120^2 - 2 * b * 120 Cos28`

`4900 = b^2 + 14400 - 2 * b * 120 * 0.8829`

`4900 = b^2 + 14400 - 211.9b`

Subtract 4900 from both sides

`b^2 + 14400 - 4900 - 211.9b = 0`

`b^2 + 9500 - 211.9b = 0`

`b^2 - 211.9b + 9500 = 0`

Solve using quadratic formula

`(-b\±√(b^2 - 4ac))/(2a)`

Substitute 1 for a; -211.9 for b and 9500 for c

`b = (-(211.9)\±√((211.9)^2 - 4 * 1 * 9500))/(2 * 1)`

`b = (211.9\±√(44901.61 - 38000))/(2)`

`b = (211.9\±√(6901.61))/(2)`

`b = (211.9\±83.08)/(2)`

This can be splitted to

`b = (211.9+83.08)/(2)` or
`b = (211.9-83.08)/(2)`

`b = (294.98)/(2)` or
`b = (128.82)/(2)`

`b = 147.49` or
`b = 64.41`

`b = 147.5km\ or\ b = 64.4km`

Hence, the two possible distance are 147.5km and 64.4km