Two boats leave a dock together. They travel in straight courses, with an angle of 136° between their courses. One boat travels at 48 mph, and other at 40 mph.

Question

How far apart are the boats after 45 minutes?

Answer 1

about 61.02 miles

45 min = 3/4 of an hour

3/4 x 48 =36 miles

3/4 x 40 = 30 miles

136 = almost 135 where

d^2 = (15sqr2)^2 + (15sqr2 + 36)^2

=450 + 3273.35 = 3723.35

d = 61.02

or law of sines

30/sin20.3 = 36/sin23.7 = d/sin136

= about 87 = d/.7

d = about .7(87= about 61 miles apart after 45 minutes

or law of cosines

d^2 = 30^2 + 36^2 -2(30)(36)cos136

=900 + 1296 - 2160(.7)

=3708

d=60.9

if 3 methods are that close, after rounding errors, 61 miles looks about right, or

keep track of more decimal places and plug it into a calculator for a more

significant decimals.

Answer 2

Use the law of cosines to solve this problem.

The first boat, call it boat A, travels at 48 mph for 45 minutes, which is three fourths of an hour, so it travels (3/4)(48)=36 miles. The second boat, boat B, travels 30 miles in the same amount of time.

Set up triangle ABC, where angle A is across from side a=36 miles, angle B is across from side b=30 miles, and angle C=138^o, across from side c, representing the distance between the two boats after 45 minutes.

Then, the law of cosines tells us that: c² = a² + b² - 2abcos C.

Plugging in the values we have, c² = 36² + 30² - 2(36)(30)cos 138^o.

Multiply to get c² = 2196 - 2160cos138^o.

Using your calculator *in degree mode*, type that expression in to get c² = 3801.192823...

Without rounding, take the square root of c² by hitting the square root key, then 2nd, (-), which is "ANS", or answer.

c = about 61.654 miles lie between the boats after 45 minutes.

I hope this helps! Trig is awesome!

Let me know if you have any more questions...

Liz Z.