Law of Sines problems that involve a perpendicular

AI Thread Summary
The discussion focuses on solving Law of Sines problems that involve perpendiculars, highlighting confusion around applying the correct relationships. For the first problem, the width of the stream is calculated using the sine function, yielding an approximate width of 400.74 feet, which differs from the book's answer of 345.43 feet. The second problem involves finding the length of AM and angles BAM and MAC, with a calculated length of 490.75 feet, close to the book's answer of 490.83 feet, suggesting a minor rounding difference. The participant acknowledges a previous error in using cosine instead of sine for the first problem, leading to incorrect results. Overall, the discussion emphasizes the importance of correctly applying trigonometric functions in these scenarios.
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Could someone point me in the correct direction? I have no problem working out the angles and lines, but when one has to take into account the perpendicular, then I get confused. It is clear that the relationships are altered, but I am missing something? I have made worked further in the problem set but these two have me a bit mixed up. These are from a website, and as I have related I am trying to re-learn old math habits. Not a "homework" question, per se.

Homework Statement



AB is a line 652 feet long on one bank of a stream, and C is a point on the opposite bank. A = 53° 18', and B = 48° 36'. Find the width of the stream from C to AB.

In a triangle ABC, a = 700 feet, B = 73° 48', and C = 37° 21'. If M is the middle point of BC find the length of AM, and the angles BAM and MAC.

Homework Equations



SinA/a = SinB/b = SinC/c

The Attempt at a Solution



sin 53.3 (652)/sin 78.1 = a, the calculator claims 534.24'
then, cos 48.6 (534.24') = c, calculator reads 353.30' WRONG!

For the second one I worked to this point:

b = sin 73.8 (700')/sin 68.85 = 720.76', then sin 37.35 (720.76) = AM. . .I saw this wasn't correct. . .
 
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For the first problem, I agree that a = 534.24', but I disagree with your final value for the width of the stream. The relationship is sin(48.6 deg) = w/534.24, so w = 534.24 * sin(48.6 deg) = 400.74' (approx).
 
For the second problem, after you have found b, you know MC and angle C, so you can use the law of cosines to find AM.
 
Eureka!

As for the second problem, I now get 490.75'. The answer given is 490.83'. I am thinking that is rounding difference. On the first problem, I have no idea why I used cos instead of sine. . . looking at too many problems, I suppose. However, the given answer is 345.43'. After I correct for my function error, I get the same figure that you do.
 
I don't see how they get 345.43' for the first problem. I worked it again and got 400.74' again. It's possible that the book's answer is wrong. I don't automatically assume that when my answer is different from the book's answer, but answers in books are wrong from time to time.
 
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