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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.

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.

SinA/a = SinB/b = SinC/c

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. . .

## 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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