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Need help with triangle.  What is the value of x?

 

 Jan 13, 2024

Best Answer 

 #1
avatar+287 
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By the Exterior Triangle Angle Sum Theorem ( kinda forgot name) , angle ABC + angle BAC = angle ACD, so

 

\(x^2-4x-3=9x+27\)

 

\(x^2-13x-30=0\)

 

I will use the quadratic formula to find the roots of x, and then eliminate the value of x that produces negative angle measures.

 

a = 1, b = -13, c = -30

 

\(x = {13 \pm \sqrt{169+120} \over 2}\)

 

\(x = {13 \pm 17 \over 2}\)

 

We get: x = 15, and x = -2.

 

Eliminate x = -2, and the answer is x = 15

 

Answer: 15

 Jan 13, 2024
 #1
avatar+287 
+1
Best Answer

By the Exterior Triangle Angle Sum Theorem ( kinda forgot name) , angle ABC + angle BAC = angle ACD, so

 

\(x^2-4x-3=9x+27\)

 

\(x^2-13x-30=0\)

 

I will use the quadratic formula to find the roots of x, and then eliminate the value of x that produces negative angle measures.

 

a = 1, b = -13, c = -30

 

\(x = {13 \pm \sqrt{169+120} \over 2}\)

 

\(x = {13 \pm 17 \over 2}\)

 

We get: x = 15, and x = -2.

 

Eliminate x = -2, and the answer is x = 15

 

Answer: 15

DS2011 Jan 13, 2024

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