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Geometry 7 Online
OpenStudy (anonymous):

Using the defects method, which of these relationships represents the Law of Cosines if the measure of the included angle between the sides a and b of ∆ABC is more than 90°?

OpenStudy (anonymous):

area of square c2 = -area of square a2 – area of square b2 + area of defect1 + area of defect2 area of square c2 = area of square a2 + area of square b2 + area of defect1 – area of defect2 area of square c2 = area of square a2 + area of square b2 – area of defect1 – area of defect2 area of square c2 = area of square a2 + area of square b2 + area of defect1+ area of defect2 area of square c2 = area of square a2 - area of square b2 + area of defect1 - area of defect2

OpenStudy (tygrr321):

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OpenStudy (anonymous):

wich do you think it is?

OpenStudy (anonymous):

Honestly, I have no clue what their even asking.

OpenStudy (anonymous):

hmm

OpenStudy (anonymous):

I basically need step by step help on Geometry because I literally cant comprehend what to do.

OpenStudy (anonymous):

I am not sure about the "defects method", perhaps if you could post the original term (assuming it is in a different language than English) it would be a little easier. However, to find out if the included angle of a triangle is greater than 90 degrees, we need to use the cosine rule if the measures of the three sides are known. With the usual conventions of cosine rule, where angle A is opposite side of length a, etc, we calculate cosA=b2+c2−a22bc If cos A is negative, angle A is greater than 90 degrees. This test is useful when working with the sine rule, which gives ambiguous results under certain circumstances (e.g. when only one angle is known or given)

OpenStudy (anonymous):

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OpenStudy (anonymous):

@LilliBelle

OpenStudy (anonymous):

There you go.:D

OpenStudy (anonymous):

Okay so I'm thinking it would be A then

OpenStudy (anonymous):

I cant tell you the answer:P

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