Question

Alan and Ann are standing on a riverbank, 200 meters apart, at points A and B respectively. They want to know the distance from Ann to a house located across the river at point C. Alan measures angle A (angle BAC) to be 43°, and Ann measures angle B (angle ABC) to be 61°. What is the distance from Ann to the house? Round your answer to the nearest tenth of a meter.

Answer 1

To determine the distance from Ann to the house, we use the Law of Sines with the given angles and the distance between Alan and Ann. After calculating, we find that the distance from Ann to the house is approximately 180.9 meters.

Step-by-step explanation:

To calculate the distance from Ann to the house across the river (point C), we can use the properties of a triangle formed by points A, B, and C. Given that Alan and Ann are 200 meters apart and angles A and B are 43° and 61° respectively, we can determine the distance from point B to point C using the Law of Sines in a non-right-angled triangle:

AC / sin(㥅) = BC / sin(㥃) = AB / sin(㥂)
Where:

• AC is the distance from Alan (point A) to the house (point C).
• BC is the distance from Ann (point B) to the house (point C) (what we need to find).
• AB is the distance between Alan and Ann, which is 200 meters.
• Angle 㥂 is the angle at point C opposite to AB.

First, we can find angle 㥂 since the sum of angles in any triangle is 180°:

Angle 㥂 = 180° - 43° - 61° = 76°

Next, we can apply the Law of Sines to solve for BC:

200 meters / sin(76°) = BC / sin(61°)

BC = (200 meters * sin(61°)) / sin(76°)

After computing, we get that BC ≈ 180.9 meters.

So, the distance from Ann to the house across the river is approximately 180.9 meters, which can be rounded to the nearest tenth of a meter.