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November 23, 2014

November 23, 2014

Posted by **Kaytee** on Monday, July 10, 2006 at 7:29pm.

Kaytee

If you have all the angles, and two sides..you can use the law of sines, or the law of cosines.

Law of sines:

a/SinA = b/SinB = c/sinC

law of cosines:

c^2= a^2 + b^2 -2abCosC

The law of sines will be easier.

Thank you but I only have one side (the height) and all 3 angles. Any other solution?

Give us those angles and side and height

Use the law of sines>

a/SinA=b/sinB

You have angles A,B and side a.

Well, thank you for your insight but I was just wondering if there was a simple law that would give me a side from 2 corresponding angles. I was sure that the angles could be added in a certain fashion or something to give me the corresponding side. Unfortunately for this exam we cannot use a calculator so I have to be able to find the answer free hand; and to be honest I have never learned about the law of sines at all. I would appreciate anymore ideas. Here are the angles:

it is a right triangle with the other two angles being 30 degrees and 60 degrees with a height of 5. What I need to know is which is greater, the area of the triangle or 25*the square root of 3. In order for me to find the area though I must be able to find the base. Please help.

Kaytee, the angles you gave us are termed "nice" angles and you're expected to know the sin, cos, and tan values for angles with degree 30,45,and 60. I'll let you look up their values, but in trig we're expected to learn these ones by rote. You should see easily why they're "nice".

You might want to start a new post so we can see it; looks like this one's getting ready to go over the edge.

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