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.
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
1 answer
1 answer