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I have two questions
two forest fire towers, A and B are 20.3km apart. From tower A, the bearing of tower B is 70 degrees. The ranger in each tower observes a fire and radios the bearing from the tower. the bearing from tower A is 25 degrees and from tower B is 345 degrees. How far, to the nearest tenth of a kilometre is the fire from each tower?

the other questions is:
The interior angles of a triangle are 120 degrees, 40 degrees, and 20 degrees. The longest side is 10cm longer than the shortest side. Determine the perimeter of the triangle to the nearest centimetre.

Thanks

Never attempt a question like this without a sketch or diagram.
I labeled the position of the fire as F
and by some simple adding/subtracting of angles, I had angle A = 45° and angle B = 95°, thus angle F = 40° , and AB = 20.3

By sine law:
AF/sin95 = 20.3/sin40
AF = 20.3sin95/sin40 = appr31.46 km

BF/sin45 = 20.3/sin40
.....

the second one is quite easy,
make a sketch of the triangle, place x as the side opposite the 20° angle and (x+10) opposite the 120° angle.

by sine law :
x/sin20 = (x+10)/sin120
xsin120 = xsin20 + 10sin20
xsin120 - xsin20 = 10sin20
x(sin120 - sin20) = 10sin20
x = 10sin20/(sin120-sin20) = appr 6.527

so the smallest side is 6.527,
the largest side is 16.527

Use the sine law once more to find the third side, then add up the 3 sides.