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

Answer:

The two possible distance are 147.5km and 64.4km

Step-by-step 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

$\frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

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

$b = \frac{-(211.9)\±\sqrt{(211.9)^2 - 4 * 1 * 9500}}{2 * 1}$

$b = \frac{211.9\±\sqrt{44901.61 - 38000}}{2}$

$b = \frac{211.9\±\sqrt{6901.61}}{2}$

$b = \frac{211.9\±83.08}{2}$

This can be splitted to

$b = \frac{211.9+83.08}{2}$  or    $b = \frac{211.9-83.08}{2}$

$b = \frac{294.98}{2}$   or  $b = \frac{128.82}{2}$

$b = 147.49$   or   $b = 64.41$

$b = 147.5km\ or\ b = 64.4km$

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