Posted by **Tareena-- check my answer please** on Saturday, May 24, 2008 at 4:26pm.

The Captain of a freighter 8km fromthe nearer of two unloading docks on the shore finds that the angle between the lines of sight to the docks is 35 degrees. if the docks are 10km apart, how far is the tanker from the farther dock?

My answers :

idont know whis one is right

method # 1:

suppose b= 8, c=10,a=?

a^2+b^2=c^2

a^2=(10)^2-(8)^2=

100-64=36

taking sqaur root of 36

a=6

method #2

A= unknown

B= unknown

C=35 degrees

a=unkown

b=8km

c=10km

using law of sin

sin 35/10=sinB/8

=.45

B=.45 dgrees

then B+C-180=A

.45+35-180=144.5

A=144.5

using law of sin

sin 35/10=sin 144.5/a

=10.12

a=10.12

- Math -
**drwls**, Saturday, May 24, 2008 at 5:19pm
The law of sines is the way to do this. That is your method #2. Method #1 only applies to right triangles.

You made some algebra errors, however.

sin B = 0.4589

B = 27.3 degrees

180 - B - C = A , not what you wrote.

- Math -
**Tareena-- check my answer please**, Saturday, May 24, 2008 at 5:22pm
how does B= 27.3 how did u come up with that number

- Math -
**Tareena-- drwls check my answer please**, Saturday, May 24, 2008 at 5:32pm
thanks

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