Question

#6.) $\triangle stv \cong \triangle wqr$, the $m\angle s = (x^2 - 8)\degree$, the $m\angle r = (-8x + 1)\degree$, and the $m\angle w = (-7x)\degree$. what is the $m\angle q$?
(a) $73\degree$
(b) $59\degree$
(c) $56\degree$
(d) $44\degree$

#7.) $\triangle abc \cong \triangle dec$ by the angle - side - angle triangle congruence postulate. which pair of angles must be congruent?
(a) $\angle c \cong \angle b$
(b) $\angle c \cong \angle a$
(c) $\angle d \cong \angle b$
(d) $\angle d \cong \angle a$

#8.) based on the diagram, which of the following is always true?
(a) $\angle olm$ and $\angle pnm$ are right angles
(b) $\angle oml \cong \angle pmn$ because vertical angles are congruent
(c) $m$ is the midpoint of $\overline{ln}$
(d) $\triangle olm$ is a right triangle

#9.) given the figure below, which of the following below is true?
(a) $\triangle abc \cong \triangle dec$ by $sas$
(b) $\triangle abc \cong \triangle edc$ by $sas$
(c) $\triangle abc \cong \triangle dec$ by $aas$
(d) $\triangle abc \cong \triangle edc$ by $aas$

Explanation:

Question 6

Step1: Use Congruent Triangles Property

Since \(\triangle STV \cong \triangle WQR\), corresponding angles are equal. So \(\angle S = \angle W\), \(\angle T = \angle Q\), \(\angle V = \angle R\).
Set \(\angle S=\angle W\): \(x^{2}-8=-7x\)

Step2: Solve Quadratic Equation

Rearrange: \(x^{2}+7x - 8 = 0\)
Factor: \((x + 8)(x - 1)=0\)
Solutions: \(x=-8\) or \(x = 1\)

Step3: Check Valid Angle Measure

For \(x=-8\), \(\angle R=-8(-8)+1 = 65^{\circ}\), \(\angle W=-7(-8)=56^{\circ}\), \(\angle S=(-8)^{2}-8 = 56^{\circ}\) (valid).
For \(x = 1\), \(\angle R=-8(1)+1=-7^{\circ}\) (invalid, angle can't be negative). So \(x=-8\) is invalid? Wait, no, \(x = 1\): \(\angle R=-8(1)+1=-7\) (invalid), \(x=-8\): \(\angle R=65\), \(\angle W=56\), \(\angle S=56\). Wait, sum of angles in triangle: \(56 + \angle T+65 = 180\), \(\angle T=59\). Wait, no, \(\angle T=\angle Q\). Wait, maybe miscalculation. Wait, \(x^{2}-8=-7x\) → \(x^{2}+7x - 8 = 0\) → roots \(x = 1\) (since \(1 + 7 - 8 = 0\)) and \(x=-8\). For \(x = 1\): \(\angle S=1 - 8=-7\) (invalid). For \(x=-8\): \(\angle S=64 - 8 = 56\), \(\angle W=56\), \(\angle R=-8(-8)+1=65\). Then sum of angles in \(\triangle WQR\): \(56+\angle Q + 65=180\) → \(\angle Q=59^{\circ}\)? Wait, no, \(\angle T=\angle Q\), and in \(\triangle STV\), angles are \(\angle S = 56\), \(\angle V=\angle R = 65\), so \(\angle T=180 - 56 - 65 = 59\), so \(\angle Q = 59^{\circ}\).

Given \(\triangle ABC \cong \triangle DEC\) by ASA. Corresponding angles: \(\angle A\) corresponds to \(\angle D\), \(\angle B\) to \(\angle E\), \(\angle C\) to \(\angle C\) (common). Wait, ASA: two angles and included side. So \(\angle A\) and \(\angle B\) with included \(BC\), and \(\angle D\) and \(\angle E\) with included \(EC\). Wait, the congruence is \(\triangle ABC \cong \triangle DEC\), so \(\angle A\cong\angle D\), \(\angle B\cong\angle E\), \(\angle C\cong\angle C\). Wait, the options: (D) \(\angle D\cong\angle A\) is correct as corresponding angles.

• Option A: No info to say they are right angles.
• Option B: \(\angle OML\) and \(\angle PMN\) are vertical angles, so they are congruent (vertical angles theorem).
• Option C: No info \(M\) is midpoint.
• Option D: No info \(\triangle OLM\) is right.

So B is correct.

Answer:

B. \(59^{\circ}\)

Question 7