SOLUTION: A ship sees a lighthouse which is found to be 9.8 km distant through the use of radar. The lighthouse is 3km from a second lighthouse. The angle between the line of sight between t

Algebra ->  Triangles -> SOLUTION: A ship sees a lighthouse which is found to be 9.8 km distant through the use of radar. The lighthouse is 3km from a second lighthouse. The angle between the line of sight between t      Log On


   



Question 1189682: A ship sees a lighthouse which is found to be 9.8 km distant through the use of radar. The lighthouse is 3km from a second lighthouse. The angle between the line of sight between the two lighthouses is 30°. What is the angle between the line of sight between the ship and lighthouse 2, when looking from lighthouse 1?
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Drawing

A = ship's location
B = lighthouse 1
C = lighthouse 2

Angle A = 30 degrees
Angle B = x degrees
Angle C = 180-A-B = 180-30-x = 150-x degrees

Apply the law of sines to find x
sin(A)/a = sin(C)/c
sin(30)/3 = sin(150-x)/9.8
0.5/3 = sin(150-x)/9.8
0.5*9.8 = 3sin(150-x)
4.9 = 3sin(150-x)
3*sin(150-x) = 4.9
sin(150-x) = 1.633333 approximately

We run into a problem. Recall that -1+%3C=+sin%28x%29+%3C=+1, i.e. the outputs of sine are restricted to the interval between -1 and 1.
So it's impossible to have sin(150-x) be 1.63 as that's larger than 1.

No triangle is possible with those specified side lengths and angle.
Please contact your teacher for clarification.