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 October 14th, 2010, 07:52 PM #1 Member   Joined: Sep 2010 Posts: 37 Thanks: 0 trigonometry radians Triangle drawn inside a circle using the diameter of the circle as one side with the other 2 sides meeting at a point on the circumference of the circle. If one of the angles of the triangle is 0.95 radians and the length of the adjacent side is 10units, what is the radius of the circle? How? Also, how many radians are there in a triangle? River running parallel to the base of a cliff has a bridge over it which is 4 m longer than the river is wide. From the top of a 60m high cliff the angle of depression is 0.71 radians for one end of the bridge and 1.08 radians for the other end. How wide is the river? I don't even get how to draw the diagram for that question. Thanks.
 October 14th, 2010, 08:45 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: trigonometry radians Use the fact that the hypotenuse of a right triangle is a diameter of the triangle's circumcircle. The angle made where the shorter legs of the triangle meet on the circle is a right angle, meaning the third angle is complementary to the given angle of 0.95 radians, or ?/2 - 0.95. Then use the law of sines to state: sin(?/2)/2r = sin(?/2 - 0.95)/10 Use sin(?/2 - x) = cos(x), and multiply through by 2. 1/r = cos(0.95)/5 Invert both sides. r = 5/cos(0.95) ? 8.596 The sum of the interior angles in a triangle is ? radians.
October 14th, 2010, 09:32 PM   #3
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Re: trigonometry radians

[attachment=0vfwh5vu]tanhelp.JPG[/attachmentvfwh5vu]

w = width of river, giving

x + y = 4 m.
tan(?/2 - 0.71) = (w + x)/60
tan(?/2 - 1.0 = (w + y)/60

Can you go from there?
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October 15th, 2010, 12:04 AM   #4
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Re: trigonometry radians

You can simplify further using the identity tan(?/2 - ?) = cot(?), giving

tan(0.71) = 60/(w + x)
tan(1.0 = 60/(w + y)

If you refer to the sketch, you'll see this makes sense. Using x + y = 4, we have

[attachment=0:2471kkdd]tanhelp.JPG[/attachment:2471kkdd]

w + x = 60/tan(0.71)
w + 4 - x = 60/tan(1.0

Adding the two equations, we have

2w + 4 = 60(cot(0.71) + cot(1.0)

Dividing through by 2, then subtracting 2 from each side gives

w = 30(cot(0.71) + cot(1.0) - 2

w ? 48.94 m.
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 tanhelp.JPG (10.2 KB, 149 views)

 October 20th, 2010, 04:22 AM #5 Member   Joined: Sep 2010 Posts: 37 Thanks: 0 Re: trigonometry radians Okay....But, the solution given is 33.7m
 October 20th, 2010, 08:44 AM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond Re: trigonometry radians 60 * tan(?/2 - 0.71) - 60 * tan(?/2 - 1.0 - 4 ? 33.74, where ?/2 - 0.71 is the angle from the base of the cliff to the outer edge of the bridge and ?/2 - 1.08 is the angle from the base of the cliff to the inner edge of the bridge.
 October 20th, 2010, 09:27 AM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: trigonometry radians Sorry, I assumed, and as usual, having a 50% chance, made the wrong assumption!
 October 21st, 2010, 01:46 AM #8 Member   Joined: Sep 2010 Posts: 37 Thanks: 0 Re: trigonometry radians Could you draw the diagram ? I'm a bit confused still. THanks
 October 21st, 2010, 06:47 AM #9 Global Moderator   Joined: Dec 2006 Posts: 20,104 Thanks: 1907 You need to assume that the measurements are made from a position (on the clifftop) in the same vertical plane as the ends of the bridge, then use a diagram which shows two lines joining the ends of the bridge to a single point on the clifftop at a height of 60m above the level of the ends of the bridge. The ends of the bridge are then 60cot(0.71)m and 60cot(1.0m measured horizontally from the observation point, and so these values differ by the distance between the ends of the bridge.

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