group c - angles in parallel lines, including triangles
calculate the size of angle $\theta$. you must use the sum of angles in a triangle in each solution.
state any other angle facts that you use.

1)
$\theta$, $74^\circ$, $28^\circ$

2)
$52^\circ$, $68^\circ$, $\theta$

3)
$41^\circ$, $\theta$

4)
$32^\circ$, $116^\circ$, $\theta$

5)
$47^\circ$, $18^\circ$, $\theta$

6)
$77^\circ$, $36^\circ$, $\theta$

7)
$60^\circ$, $120^\circ$, $\theta$

8)
$135^\circ$, $\theta$

9)
$65^\circ$, $45^\circ$, $45^\circ$, $\theta$

Question

group c - angles in parallel lines, including triangles
calculate the size of angle $\theta$. you must use the sum of angles in a triangle in each solution.
state any other angle facts that you use.

1)
$\theta$, $74^\circ$, $28^\circ$

2)
$52^\circ$, $68^\circ$, $\theta$

3)
$41^\circ$, $\theta$

4)
$32^\circ$, $116^\circ$, $\theta$

5)
$47^\circ$, $18^\circ$, $\theta$

6)
$77^\circ$, $36^\circ$, $\theta$

7)
$60^\circ$, $120^\circ$, $\theta$

8)
$135^\circ$, $\theta$

9)
$65^\circ$, $45^\circ$, $45^\circ$, $\theta$

Accepted Answer

{ "quality": "clear", "question_count": 9, "skills_matched": [ "step_cot", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Parallel Line Angles", "Transversal Lines", "Angle Relationships" ], "new_concepts": [ "Triangle Angle Sum Theorem" ], "current_concepts": [ "Parallel Line Angles", "Transversal Lines", "Angle Relationships", "Triangle Angle Sum Theorem" ] } ## Solve Question 1 Using the *Parallel Line Angles* and *Angle Relationships* knowledge points $$ \begin{aligned} &\angle ECD = 180^\circ - 74^\circ = 106^\circ \quad \text{(Angles on a straight line)} \ &\angle CED = 180^\circ - (106^\circ + 28^\circ) = 46^\circ \quad \text{(Sum of angles in }\triangle ECD\text{)} \ &\angle AEB = \angle CED = 46^\circ \quad \text{(Vertically opposite angles)} \ &\theta = \angle AEB = 46^\circ \quad \text{(Corresponding angles, } AB \parallel EC\text{)} \end{aligned} $$ ## Solve Question 2 Using the *Parallel Line Angles* and *Angle Relationships* knowledge points $$ \begin{aligned} &\angle BAC = 52^\circ \quad \text{(Alternate interior angles, } AB \parallel \text{bottom line)} \ &\angle BCA = 180^\circ - (52^\circ + 68^\circ) = 60^\circ \quad \text{(Sum of angles in }\triangle ABC\text{)} \ &\angle BCD = 180^\circ - 60^\circ = 120^\circ \quad \text{(Angles on a straight line)} \ &\theta = 180^\circ - 120^\circ = 60^\circ \quad \text{(Interior angles on the same side, } AB \parallel \text{bottom line)} \end{aligned} $$ ## Solve Question 3 Using the *Parallel Line Angles* and *Angle Relationships* knowledge points $$ \begin{aligned} &\triangle ABC \text{ is isosceles with } AC = AB \implies \angle ACB = \angle ABC = 41^\circ \ &\angle BAC = 180^\circ - (41^\circ + 41^\circ) = 98^\circ \quad \text{(Sum of angles in }\triangle ABC\text{)} \ &\angle CDE = \angle BAC = 98^\circ \quad \text{(Alternate interior angles, } AB \parallel ED\text{)} \ &\theta = 180^\circ - 98^\circ = 82^\circ \quad \text{(Angles on a straight line)} \end{aligned} $$ ## Solve Question 4 Using the *Parallel Line Angles* and *Angle Relationships* knowledge points $$ \begin{aligned} &\angle BCD = 180^\circ - 116^\circ = 64^\circ \quad \text{(Angles on a straight line)} \ &\triangle ABD \text{ is isosceles with } AB = BD \implies \angle ADB = \angle DAB = 32^\circ \ &\angle CBD = 32^\circ + 32^\circ = 64^\circ \quad \text{(Exterior angle of }\triangle ABD\text{)} \ &\angle BDC = 180^\circ - (64^\circ + 64^\circ) = 52^\circ \quad \text{(Sum of angles in }\triangle BCD\text{)} \ &\theta = 180^\circ - (32^\circ + 52^\circ) = 96^\circ \quad \text{(Angles on a straight line)} \end{aligned} $$ ## Solve Question 5 We need to find $\theta$ in Question 5. First, let's identify the angles in $\triangle ABC$. The top line containing $B$ is parallel to the bottom line containing $A$ and $C$. Thus, the alternate interior angle to $\angle ABC$ is $\angle BAA'$ (where $A'$ is to the left on the bottom line). Specifically, $\angle BAC$ and the angle of $47^\circ + 18^\circ = 65^\circ$ at $B$ are alternate interior angles: $$ \angle BAC = 65^\circ $$ In $\triangle ABC$, the sum of angles is $180^\circ$: $$ \angle BCA = 180^\circ - (\angle BAC + \angle ABC) $$ Since $\angle ABC = 47^\circ$: $$ \angle BCA = 180^\circ - (65^\circ + 47^\circ) = 68^\circ $$ Now, looking at the triangle containing $\theta$, which is $\triangle PAC$ (where $P$ is the point on the segment $AB$): The angle $\theta$ is an exterior angle or part of a triangle. Let's use the angle sum of $\triangle AB'C$ where the line segment from $B$ goes to the bottom line at an angle of $18^\circ$ to the vertical-like line. The angle at the bottom left of this inner triangle is $\theta$. Using the sum of angles in the inner triangle: $$ \theta = 180^\circ - (18^\circ + 68^\circ) = 94^\circ $$ ## Question 1 $$\theta = 46^\circ$$ ## Question 2 $$\theta = 60^\circ$$ ## Question 3 $$\theta = 82^\circ$$ ## Question 4 $$\theta = 96^\circ$$ ## Question 5 $$\theta = 94^\circ$$ ## Question 6 $$\theta = 113^\circ$$ ## Question 7 $$\theta = 60^\circ$$ ## Question 8 $$\theta = 45^\circ$$ ## Question 9 $$\theta = 110^\circ$$ { "subject": "Mathematics", "question_type": "Multi-part", "knowledge_point": [ "Mathematics", "Geometry", "Angle Relationships" ] }

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