Question

A surveyor starts at the edge of a lake, at point A, and measures 90° angle between B and C, as shown. The surveyor then walks 85 meters to B and measures an angle of 62° between A and C. Approximately how many meters wide is the lake between A and C? Round your answer to the nearest hundredth if necessary.

Answer 1

Answer: We can use the Law of Sines to solve for the distance between points A and C. Let x represent the distance between A and C in meters. Then, we have:

sin(62°) = x / sin(28°)

and

sin(90°) = 85 / sin(28°)

Simplifying the second equation, we get:

sin(28°) = 85 / cos(90°)

sin(28°) = 85 / 0

This is undefined, which means that our assumption that angle C is acute (less than 90°) was incorrect. Instead, we know that angle C must be obtuse (greater than 90°). To find the correct value of angle C, we can use the fact that the three angles of a triangle must add up to 180°:

angle A + angle B + angle C = 180°

90° + 62° + angle C = 180°

angle C = 28°

Now we can use the Law of Sines as before:

sin(62°) = x / sin(28°)

x = sin(28°) * (85 / sin(62°))

x ≈ 69.57

Rounding to the nearest hundredth, we get:

x ≈ 69.57 meters

Therefore, the lake is approximately 69.57 meters wide between points A and C.