Question 1122327: The ship S is sighted from two lighthouses B and U which are 16 km apart. If m∠SBU= 27° and m∠SUB= 65°. Find the distance from S to the nearer lighthouse.
Found 2 solutions by Boreal, Theo:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Draw this out. The nearest lighthouse is U as SU is opposite the smallest angle, SBU, in the triangle.
Law of Sines: the distance between the two lighthouses, 16 km, is opposite to the angle BSU, which is 88 degrees, since we know the other two angles of the triangle.
16/sin 88=x/sin 27
x=16*sin 27/sin 88
x=7.27 km
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! this forms triangle SBU where angle SBU = 27 degrees and angle SUB = 65 degrees.
angle BSU is equal to 180 - 27 - 65 = 88 degrees.
88 + 27 + 65 = 180, as it should, since the sum of the interior angles of a triangle is 180 degrees.
BU = 16 kilometers.
use the law of sines to find the length of sides BS and SU.
law of sines says: a/sin(A) = b/sin(B) = c/sin(C)
applied to this triangle, you get 16/sin(88) = SU/sin(27) = SB/sin(65)
this can be broken up into:
16/sin(88) = SU/sin(27) and 16/sin(88) = SB/sin(65)
cross multiply each to get:
16 * sin(27) = sin(88) * SU and 16 * sin(65) = SB * sin(88)
solve for SU to get SU = 16 * sin(27) / sin(88) = 7.298275633
solve for SB to get SB = 16 * sin(65) / sin(88) = 14.50976355
the distance from S to the nearer lighthouse is 7.298275633 kilometers.
|
|
|
