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Question 1042290: Law of Sines: Triangle question. What is the Distance from point A to point C, given two angles and one length?
Text of the problem reads:
Maya and Juan are standing on a riverbank, 230 meters apart, at points A and B respectively. (See figure). They want to know the distance from Maya (point A) to a house (point C). Maya (point A) measures angle A, angle BAC to be 48 degrees, Juan (point B) measures angle B, angle ABC to be 62 degrees. What is the distance from Maya (point A) to the house (point C)? Round your answer to the nearest tenth of a meter.
I have a .png to upload, but not sure how to upload the .png image.
Precalc_23.png
Since I cannot upload the image: here is a description to draw a diagram:
Draw a triangle and label the vertices A, B and C.
Label point A Maya
Label point B Juan
Label point C the house
Label angle BAC as 48 degrees.
Label angle ABC as 62 degrees.
Label side AB as 230 m
Label side AC as x (since that is what we are looking to find).
I am thinking the easiet way to find x is to use the Law of Sines. But I am not sure how to set up the equation for Law of Sines.
I know it will require that we know what the angle opposite side AB is. Using the fact that the three angles of a triangle add up to 180 degrees we should be able to use the two given angles to find the third one. The third angle, ACB, is 70 degrees.
The unknown angle ACB is 70 degrees: determined because we know a triangle is 180 degrees total. Therefore the given angles of 48 + 62 = 110, gives us the unknown angle located at point C. (Point C is the house in the diagram). 180 - 110 = 70 degrees for point C, (angle ACB). The question is asking what the distance is from point A (Maya) to point C (the house). Distance between point A and B is 230 m. Two sides of the triangle are not labeled with a distance.
Help is appreciated to complete the problem and answer What is the distance from Maya (point A) to the house (point C)?
Answer by josgarithmetic(39616)   (Show Source):
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