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February 10th, 2016, 01:28 AM   #1
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Worded question

Worded question that I have no idea how to answer -

Points A B and C all lie on the same horizontal ground with B due north of A and C on a bearing of 030* from A.
From the top of a vertical tower at A, 37 metres above ground, point B has an angle of depression of 17* and point C has an angle of depression of 12*.
How far is B from C?

Last edited by skipjack; February 16th, 2019 at 04:02 AM.
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February 10th, 2016, 05:01 AM   #2
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There are three triangles of importance here. One is the right triangle formed by the tower at A, the line from A to B, and the line of sight from the top of the tower to B. If we let "x" be the distance from A to B then x/37= tan(17). That's easy to solve for x.

Another is the right triangle formed by the tower at A, the line from A to C, and the line of sight from the top of the tower to C. If we let "y" be the distance from A to C, then y/37= tan(12). That's easy to solve for y.

The third triangle is the right triangle formed by the lines from A to B, from A to C, and from B to C. Call the distance from B to C "z". By the "cosine law" $\displaystyle z^2= x^2+ y^2- 2xy \cos(30^\circ)$.

Last edited by skipjack; February 16th, 2019 at 04:04 AM.
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February 12th, 2016, 12:39 AM   #3
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Thank you. :d

Last edited by skipjack; February 16th, 2019 at 04:05 AM.
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February 15th, 2019, 07:30 PM   #4
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Can someone please draw the diagram for that question and post it here?
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February 15th, 2019, 10:53 PM   #5
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Quote:
Originally Posted by koolie View Post
Can someone please draw the diagram for that question and post it here?
You want him to search the diagram for a question he asked 3 years ago to post it here for no reason?
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February 16th, 2019, 04:15 AM   #6
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The third triangle that Country Boy described doesn't have a right angle, but is otherwise correctly described.
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