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Answer: D) 130.5 degrees</h3>
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Work Shown:
![c^2 = a^2 + b^2 - 2*a*b*\cos(C)\\\\10^2 = 5^2 + 6^2 - 2*5*6*\cos(x)\\\\100 = 25 + 36 - 60*\cos(x)\\\\100 = 61 - 60*\cos(x)\\\\100 - 61 = - 60*\cos(x)\\\\39 = - 60*\cos(x)\\\\\cos(x) = \frac{39}{-60}\\\\\cos(x) = -0.65\\\\x = \arccos(-0.65)\\\\x \approx 130.5416\\\\x \approx 130.5\\\\](https://tex.z-dn.net/?f=c%5E2%20%3D%20a%5E2%20%2B%20b%5E2%20-%202%2Aa%2Ab%2A%5Ccos%28C%29%5C%5C%5C%5C10%5E2%20%3D%205%5E2%20%2B%206%5E2%20-%202%2A5%2A6%2A%5Ccos%28x%29%5C%5C%5C%5C100%20%3D%2025%20%2B%2036%20-%2060%2A%5Ccos%28x%29%5C%5C%5C%5C100%20%3D%2061%20-%2060%2A%5Ccos%28x%29%5C%5C%5C%5C100%20-%2061%20%3D%20-%2060%2A%5Ccos%28x%29%5C%5C%5C%5C39%20%3D%20-%2060%2A%5Ccos%28x%29%5C%5C%5C%5C%5Ccos%28x%29%20%3D%20%5Cfrac%7B39%7D%7B-60%7D%5C%5C%5C%5C%5Ccos%28x%29%20%3D%20-0.65%5C%5C%5C%5Cx%20%3D%20%5Carccos%28-0.65%29%5C%5C%5C%5Cx%20%5Capprox%20130.5416%5C%5C%5C%5Cx%20%5Capprox%20130.5%5C%5C%5C%5C)
Note: I used the law of cosines. Make sure your calculator is in degree mode.
N - 2 = 10n + 4/2
-2 = 9n + 4/2
-4/4 = 9n
-1 = 9n
-9 = n
N is -9.
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Step-by-step explanation:
HL states that a hypotenuse and leg of a right triangle must be congruent, so you would need to know the hypotenuses are congruent and a leg is congruent. For 6, You're only given 2 sides instead of 3, and since this is a Side-Side-Side Postulate, you must have the third sides be congruent. For 7, Angle-Side-Angle requires to known congruent angles and a side <em>in between </em>them, so you would need to know that angle I is congruent to angle L. Lastly, Side-Angle-Side needs two sides and an angle between them, so you would need to know that side JI is congruent to HD. HI is already congruent to itself by the Reflexive Property, so you only need to know JI ≅ HD.