Question

14
note: figure not drawn to scale.
if ( mangle j=(x^{2}+6x + 33)^{circ},mangle k=(x^{2}+3x - 9)^{circ}), and ( mangle l=(x^{2}-9x + 129)^{circ}), which of the following could be ( mangle m)?
a. ( 120^{circ})
b. ( 69^{circ})
c. ( 60^{circ})
d. ( 15^{circ})

Explanation:

Step1: Use the exterior - angle theorem

The exterior - angle theorem states that \(m\angle L=m\angle J + m\angle K\).
Substitute the given angle expressions:
\(x^{2}-9x + 129=(x^{2}+6x + 33)+(x^{2}+3x - 9)\)

Step2: Simplify the right - hand side

\(x^{2}-9x + 129=x^{2}+6x + 33+x^{2}+3x - 9\)
\(x^{2}-9x + 129=x^{2}+x^{2}+6x + 3x+33 - 9\)
\(x^{2}-9x + 129=2x^{2}+9x + 24\)

Step3: Rearrange the equation to form a quadratic equation

Move all terms to one side:
\(2x^{2}+9x + 24-(x^{2}-9x + 129)=0\)
\(2x^{2}+9x + 24 - x^{2}+9x - 129 = 0\)
\(x^{2}+18x - 105 = 0\)

Step4: Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

For the quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 1\), \(b = 18\), \(c=-105\))
\(x=\frac{-18\pm\sqrt{18^{2}-4\times1\times(- 105)}}{2\times1}=\frac{-18\pm\sqrt{324 + 420}}{2}=\frac{-18\pm\sqrt{744}}{2}=\frac{-18\pm2\sqrt{186}}{2}=-9\pm\sqrt{186}\)

Another way:
Since \(m\angle L + m\angle M=180^{\circ}\) (linear pair), \(m\angle M = 180-(x^{2}-9x + 129)=-x^{2}+9x + 51\)

Also, from \(m\angle L=m\angle J + m\angle K\)
\(x^{2}-9x + 129=x^{2}+6x + 33+x^{2}+3x - 9\)
\(x^{2}+18x - 105 = 0\)
\((x + 21)(x - 5)=0\) (by factoring \(x^{2}+18x - 105=(x + 21)(x - 5)\) where \(21\times(-5)=-105\) and \(21+( - 5)=16\) (wrong, actually \(x=\frac{-18\pm\sqrt{324 + 420}}{2}=\frac{-18\pm\sqrt{744}}{2}\), but if we assume factoring error and use the property of angles)

Since angles must be positive.
Let's check by substituting \(x = 5\)
\(m\angle J=(5)^{2}+6\times5 + 33=25+30 + 33=88^{\circ}\)
\(m\angle K=(5)^{2}+3\times5 - 9=25 + 15-9=31^{\circ}\)
\(m\angle L=(5)^{2}-9\times5 + 129=25-45 + 129=109^{\circ}\) (wrong)

Let's use the linear - pair and angle - sum relationship correctly.
We know that \(m\angle M=180 - m\angle L\)
Also, from \(m\angle L=m\angle J + m\angle K\)
\(m\angle M=180-(m\angle J + m\angle K)\)
\(m\angle J+m\angle K=(x^{2}+6x + 33)+(x^{2}+3x - 9)=2x^{2}+9x + 24\)
\(m\angle M=180-(2x^{2}+9x + 24)=-2x^{2}-9x + 156\)

Another approach:
Since \(m\angle M = 180-(x^{2}-9x + 129)\)
If we assume \(x = 3\)
\(m\angle J=(3)^{2}+6\times3 + 33=9 + 18+33=60^{\circ}\)
\(m\angle K=(3)^{2}+3\times3 - 9=9 + 9-9=9^{\circ}\)
\(m\angle L=(3)^{2}-9\times3 + 129=9-27 + 129=111^{\circ}\) (wrong)

If \(x = 6\)
\(m\angle J=(6)^{2}+6\times6 + 33=36+36 + 33=105^{\circ}\)
\(m\angle K=(6)^{2}+3\times6 - 9=36 + 18-9=45^{\circ}\)
\(m\angle L=(6)^{2}-9\times6 + 129=36-54 + 129=111^{\circ}\) (wrong)

Let's use the property that \(m\angle M=180 - m\angle L\) and \(m\angle L=m\angle J + m\angle K\)
\(m\angle M=180-(m\angle J + m\angle K)\)
Let's check option by option.
If \(m\angle M = 69^{\circ}\), then \(m\angle L=180 - 69=111^{\circ}\)
If \(m\angle J + m\angle K=(x^{2}+6x + 33)+(x^{2}+3x - 9)=2x^{2}+9x + 24\)
Set \(2x^{2}+9x + 24 = 111\)
\(2x^{2}+9x-87 = 0\)
\(x=\frac{-9\pm\sqrt{81+696}}{4}=\frac{-9\pm\sqrt{777}}{4}\) (not nice)

If \(m\angle M = 15^{\circ}\), then \(m\angle L=180 - 15=165^{\circ}\)
\(m\angle J + m\angle K=(x^{2}+6x + 33)+(x^{2}+3x - 9)=2x^{2}+9x + 24\)
Set \(2x^{2}+9x + 24=165\)
\(2x^{2}+9x - 141 = 0\)
\(x=\frac{-9\pm\sqrt{81 + 1128}}{4}=\frac{-9\pm\sqrt{1209}}{4}\) (not nice)

If \(m\angle M=60^{\circ}\), then \(m\angle L = 120^{\circ}\)
\(m\angle J + m\angle K=(x^{2}+6x + 33)+(x^{2}+3x - 9)=2x^{2}+9x + 24\)
Set \(2x^{2}+9x + 24 = 120\)
\(2x^{2}+9x-96 = 0\)
\(x=\frac{-9\pm\sqrt{81+768}}{4}=\frac{-9\pm\sqrt{849}}{4}\) (not nice)

If \(m\angle M = 120^{\circ}\), then \(m\angle L=60^{\circ}\)
\(m\angle J + m\angle K=(x^{2}+6x + 33)+(x^{2}+3x - 9)=2x^{2}+9x + 24\)
Set \(2x^{2}+9x + 24=60\)
\(2x^{2}+9x - 36 = 0\)
\(x^{2}+\frac{9}{2}x-18 = 0\)
Using the quadratic formula \(x=\frac{-\frac{9}{2}\pm\sqrt{\frac{81}{4}+72}}{2}=\frac{-\frac{9}{2}\pm\sqrt{\frac{81 + 288}{4}}}{2}=\frac{-\frac{9}{2}\pm\frac{\sqrt{369}}{2}}{2}\)

Another way:
We know that in a triangle, the sum of interior angles is \(180^{\circ}\) and \(m\angle L=m\angle J + m\angle K\) (exterior - angle theorem) and \(m\angle L+m\angle M = 180^{\circ}\)
\(m\angle M=180-(m\angle J + m\angle K)\)
Let's assume \(x = 3\)
\(m\angle J=(3)^{2}+6\times3 + 33=9 + 18+33=60^{\circ}\)
\(m\angle K=(3)^{2}+3\times3 - 9=9 + 15 - 9=15^{\circ}\)
\(m\angle J + m\angle K=60 + 15=75^{\circ}\)
\(m\angle L=75^{\circ}\) (wrong as \(m\angle L=x^{2}-9x + 129=(3)^{2}-9\times3 + 129=9-27 + 129=111^{\circ}\))

Let's use the fact that \(m\angle M = 180-(x^{2}-9x + 129)\)
If we consider the options:
We know that \(m\angle M=180 - m\angle L\) and \(m\angle L=m\angle J + m\angle K\)
Let's check by substituting \(x = 5\) (trial and error for simplicity of calculation)
\(m\angle J=(5)^{2}+6\times5 + 33=25 + 30+33=88^{\circ}\)
\(m\angle K=(5)^{2}+3\times5 - 9=25 + 15 - 9=31^{\circ}\)
\(m\angle L=88 + 31=119^{\circ}\) (wrong)

If \(x = 4\)
\(m\angle J=(4)^{2}+6\times4 + 33=16+24 + 33=73^{\circ}\)
\(m\angle K=(4)^{2}+3\times4 - 9=16 + 12-9=19^{\circ}\)
\(m\angle L=73 + 19=92^{\circ}\)
\(m\angle M=180 - 92=88^{\circ}\) (wrong)

If \(x= 2\)
\(m\angle J=(2)^{2}+6\times2 + 33=4 + 12+33=49^{\circ}\)
\(m\angle K=(2)^{2}+3\times2 - 9=4 + 6 - 9=1^{\circ}\) (wrong as angle can't be \(1^{\circ}\) practically in this context)

Let's use the property \(m\angle M = 180-(m\angle J + m\angle K)\)
\(m\angle J + m\angle K=(x^{2}+6x + 33)+(x^{2}+3x - 9)=2x^{2}+9x + 24\)
We want to find \(x\) such that \(m\angle M\) is one of the options.
If \(m\angle M = 69^{\circ}\)
\(180-(2x^{2}+9x + 24)=69\)
\(2x^{2}+9x + 24=111\)
\(2x^{2}+9x - 87 = 0\)
\(x=\frac{-9\pm\sqrt{81+696}}{4}=\frac{-9\pm\sqrt{777}}{4}\approx\frac{-9\pm27.87}{4}\)
\(x=\frac{-9 + 27.87}{4}\approx4.72\) or \(x=\frac{-9 - 27.87}{4}\approx-9.22\) (reject negative \(x\))
\(m\angle J=(4.72)^{2}+6\times4.72 + 33\approx22.28+28.32 + 33=83.6^{\circ}\)
\(m\angle K=(4.72)^{2}+3\times4.72 - 9\approx22.28+14.16-9=27.44^{\circ}\)
\(m\angle L=83.6 + 27.44 = 111.04^{\circ}\)
\(m\angle M=180 - 111.04=68.96\approx69^{\circ}\)

Answer:

B. \(69^{\circ}\)