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# Trigonometry Help Please!

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ZZZZZZ  May 22, 2017
edited by ZZZZZZ  May 22, 2017

#1
+4174
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Here I made a triangle where A is the first sighting, B is the light, and C is the second sighting.

The question is, what is the length of BC ?

On this first one, I just focused on the angles.

m∠BAC = 107º - 60º   =   47º

measure of the green angle = 180º - 107º   =   73º

Since the two vertical lines are parallel and line AC crosses them, we have two parallel lines cut by a transversal. So.. the two green angles are the same.

Now let's just look at the second triangle.

We can use law of sines to find BC.

$$\frac{1.5}{\sin 26}=\frac{BC}{\sin 47} \\~\\ \frac{1.5}{\sin 26}*\sin 47 =BC \\~\\ 2.503\text{ km} \approx BC$$

hectictar  May 23, 2017
Sort:

#1
+4174
+2

Here I made a triangle where A is the first sighting, B is the light, and C is the second sighting.

The question is, what is the length of BC ?

On this first one, I just focused on the angles.

m∠BAC = 107º - 60º   =   47º

measure of the green angle = 180º - 107º   =   73º

Since the two vertical lines are parallel and line AC crosses them, we have two parallel lines cut by a transversal. So.. the two green angles are the same.

Now let's just look at the second triangle.

We can use law of sines to find BC.

$$\frac{1.5}{\sin 26}=\frac{BC}{\sin 47} \\~\\ \frac{1.5}{\sin 26}*\sin 47 =BC \\~\\ 2.503\text{ km} \approx BC$$

hectictar  May 23, 2017
#2
+75368
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Very nice, hectictar.......!!!

CPhill  May 23, 2017
#3
+312
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thanks hectictar

ZZZZZZ  May 30, 2017

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