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the lengths of the tree sides of a triangle are denoted a,b, and c; the angles opposiete these sides are A,B,C. use the given information to find the required quantities. a) B= 90 deg.; cotA=4/3; b=5 sqrt2; FIND a. b) b=10 cm, c=10cm, A = 60 deg. FIND area of ABC TRIANGLE. I can start if off but i dont know what to do next?

Mathematics
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|dw:1355085335738:dw|
Do you have a figure for this? If not, we can draw one.
well the picture i drew is all i have :/

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ok lets see
k
the lengths of the tree sides of a triangle are denoted a,b, and c; the angles opposiete these sides are A,B,C. use the given information to find the required quantities. a) B= 90 deg.; cotA=4/3; b=5 sqrt2; FIND a. Take the figure; note that the sides are a, b, and c. The angles opposite to these are A, B, and C. This means (look at the figure). |dw:1355085890124:dw|
I think according to the trigonometric formula, a = 3 But the Pythagoras theorem isn't satisfied for this one.
okay then, so basically the cot is 1/ tan, or the tangent just opposite so instead of oppoiste over adj. its adj/ opposite? and that is how we got the a, right?
b) b=10 cm, c=10cm, A = 60 deg. FIND area of ABC TRIANGLE.|dw:1355086191800:dw|
so then, something like. okay so 180-60 is 120, since b and c are equal u can divide by 2, so that means they are all 60 degrees right?
So the angles of triangle ABC would be 60 + x + x = 180 60 + 2x = 180 2x = 180 - 60 2x = 120 x = 120/2 = 60 So, angle B = angle C = 60 degrees. Triangle ABC is an equilateral triangle, so the area of an equilateral triangle is given by \[Area = \frac{ s ^{2}*\sqrt{3} }{ 4 } = \frac{ (10)^{2}*\sqrt{3} }{ 4 } = \frac{ 100\sqrt{3} }{ 4 }\]
If you want, take the decimal value for square root of 3 is 1.73. The final answer will then be something around 43.25 square units.
wait, how did we get the 5 squared?
@Fall12-13 where did the
@Fall12-13 where did the \[5^{2}\] come from?
or is it a 5?
You mean the 'square units'?
oh haha i see it its an s for the sides okay
Alright.

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