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# trigonometry question?

Points A and C are on the same side of a canyon. Point B is on the opposite side of the canyon. Angle BAC = 90^o. Point A is 100 meters away from C in the direction N 42.6^o W. From A, point B is in the direction N 73.5^o E. Find the width of the canyon.

this isnt that hard, i just keep going in circles trying to figure it out. ive drawn it out and for some reason cant figure out one of the other angles in triangle abc. im confused on how to use the directions for angles in the triangle. can anyone help please  We can do the following:

• Draw a horizontal line through C in East-West direction

• Draw a vertical line through A in North-South direction

• Now the two lines intersect at a point D and the angle is 90Â°

• Now the angle between CA and the line through C straight north is 42.6Â°

• This means that the angle between CA and CD is 90Â° - 42.6Â° = 47.4Â°

• It also means that the angle between AD and AC is 180Â° - 90Â° - 47.4Â° = 42.6Â°

• Since the angle between the line through A straight north and AB is 73.5Â° the angle between AB and AC has to be 180Â° - 73.5Â° - 42.6Â° = 63.9Â°

• Since the angle between BA and BC is 90Â° the angle between CA and CB is 180Â° - 63.9Â° - 90Â° = 26.1Â°

• Now with this we can calculate AB as AB = sin(26.1Â°) * AC = sin(26.1Â°) * 100m = 43.99m

It is best to draw a picture first to see all the angles.