QUESTION IMAGE
Question
in exercises 29–36, find the measure of the numbered angle.
- $\angle 1$ 30. $\angle 2$
- $\angle 3$ 32. $\angle 4$
- $\angle 5$ 34. $\angle 6$
- $\angle 7$ 36. $\angle 8$
Let's solve for the angles one by one. We'll start with $\angle 1$ (problem 29) and then move to others as needed.
Problem 29: Find $\angle 1$
The triangle containing $\angle 1$ is a right triangle (has a right angle, $90^\circ$) and another angle of $40^\circ$. The sum of angles in a triangle is $180^\circ$.
Step 1: Recall triangle angle sum
The sum of the interior angles of a triangle is $180^\circ$. For a right triangle, one angle is $90^\circ$, another is given as $40^\circ$, and $\angle 1$ is the third angle.
Step 2: Calculate $\angle 1$
Let the angles be $90^\circ$, $40^\circ$, and $\angle 1$. So:
Simplify:
Subtract $130^\circ$ from both sides:
Problem 30: Find $\angle 2$
$\angle 1$ and $\angle 2$ are vertical angles? Wait, no, actually, $\angle 1$ and $\angle 2$ are adjacent to form a linear pair? Wait, no, looking at the diagram, $\angle 1$ and $\angle 2$ are vertical angles? Wait, no, the intersection of two lines: $\angle 1$ and $\angle 3$ are vertical, $\angle 2$ and $\angle 4$ are vertical? Wait, no, let's see. Wait, $\angle 1$ and $\angle 2$ are supplementary? Wait, no, the triangle with $\angle 1$ is a right triangle, and the other triangle (below) is also a right triangle. Wait, actually, $\angle 1$ and $\angle 2$: since $\angle 1$ is $50^\circ$, and $\angle 1$ and $\angle 2$ are vertical angles? Wait, no, maybe $\angle 1$ and $\angle 2$ are supplementary? Wait, no, let's check the diagram again. Wait, the two triangles are congruent? Wait, the top triangle has a right angle, $40^\circ$, and $\angle 1 = 50^\circ$. The bottom triangle also has a right angle, so $\angle 5$ should be $40^\circ$ (since the angles in the triangles are equal). Then, $\angle 1$ and $\angle 2$: since they are vertical angles? Wait, no, the lines intersect, so $\angle 1$ and $\angle 3$ are vertical, $\angle 2$ and $\angle 4$ are vertical. Wait, maybe $\angle 1$ and $\angle 2$ are supplementary? Wait, no, let's calculate $\angle 2$. Wait, the sum of angles around a point is $360^\circ$, but at the intersection, the angles are $\angle 1$, $\angle 2$, $\angle 3$, $\angle 4$. So $\angle 1 + \angle 2 + \angle 3 + \angle 4 = 360^\circ$. But $\angle 1 = \angle 3$ (vertical angles), $\angle 2 = \angle 4$ (vertical angles). So $2\angle 1 + 2\angle 2 = 360^\circ$, so $\angle 1 + \angle 2 = 180^\circ$. So $\angle 2 = 180^\circ - \angle 1 = 180^\circ - 50^\circ = 130^\circ$? Wait, that can't be. Wait, maybe I made a mistake. Wait, the top triangle: right angle, $40^\circ$, so $\angle 1 = 50^\circ$ (correct, because $90 + 40 + 50 = 180$). Then, the bottom triangle: right angle, so $\angle 5 = 40^\circ$, so $\angle 3 = 50^\circ$ (same as $\angle 1$). Then, at the intersection, $\angle 1 = 50^\circ$, $\angle 3 = 50^\circ$, so $\angle 2 = 180 - 50 = 130^\circ$? Wait, but that seems odd. Wait, maybe $\angle 2$ is equal to $\angle 1$? No, that doesn't make sense. Wait, maybe the two triangles are similar, so $\angle 1 = \angle 3 = 50^\circ$, and $\angle 2 = \angle 4 = 130^\circ$? Wait, but let's check problem 31: $\angle 3$. $\angle 3$ should be equal to $\angle 1$ (vertical angles), so $\angle 3 = 50^\circ$. Then problem 32: $\angle 4$ should be equal to $\angle 2$ (vertical angles), so $\angle 4 = 130^\circ$. Then problem 33: $\angle 5$. The bottom triangle is a right triangle, so $\angle 5 + 90^\circ + \angle 3 = 180^\circ$. $\angle 3 = 50^\circ$, so $\angle 5 = 180 - 90 - 50 = 40^\circ$. Then problem 34:…
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Let's solve for the angles one by one. We'll start with $\angle 1$ (problem 29) and then move to others as needed.
Problem 29: Find $\angle 1$
The triangle containing $\angle 1$ is a right triangle (has a right angle, $90^\circ$) and another angle of $40^\circ$. The sum of angles in a triangle is $180^\circ$.
Step 1: Recall triangle angle sum
The sum of the interior angles of a triangle is $180^\circ$. For a right triangle, one angle is $90^\circ$, another is given as $40^\circ$, and $\angle 1$ is the third angle.
Step 2: Calculate $\angle 1$
Let the angles be $90^\circ$, $40^\circ$, and $\angle 1$. So:
Simplify:
Subtract $130^\circ$ from both sides:
Problem 30: Find $\angle 2$
$\angle 1$ and $\angle 2$ are vertical angles? Wait, no, actually, $\angle 1$ and $\angle 2$ are adjacent to form a linear pair? Wait, no, looking at the diagram, $\angle 1$ and $\angle 2$ are vertical angles? Wait, no, the intersection of two lines: $\angle 1$ and $\angle 3$ are vertical, $\angle 2$ and $\angle 4$ are vertical? Wait, no, let's see. Wait, $\angle 1$ and $\angle 2$ are supplementary? Wait, no, the triangle with $\angle 1$ is a right triangle, and the other triangle (below) is also a right triangle. Wait, actually, $\angle 1$ and $\angle 2$: since $\angle 1$ is $50^\circ$, and $\angle 1$ and $\angle 2$ are vertical angles? Wait, no, maybe $\angle 1$ and $\angle 2$ are supplementary? Wait, no, let's check the diagram again. Wait, the two triangles are congruent? Wait, the top triangle has a right angle, $40^\circ$, and $\angle 1 = 50^\circ$. The bottom triangle also has a right angle, so $\angle 5$ should be $40^\circ$ (since the angles in the triangles are equal). Then, $\angle 1$ and $\angle 2$: since they are vertical angles? Wait, no, the lines intersect, so $\angle 1$ and $\angle 3$ are vertical, $\angle 2$ and $\angle 4$ are vertical. Wait, maybe $\angle 1$ and $\angle 2$ are supplementary? Wait, no, let's calculate $\angle 2$. Wait, the sum of angles around a point is $360^\circ$, but at the intersection, the angles are $\angle 1$, $\angle 2$, $\angle 3$, $\angle 4$. So $\angle 1 + \angle 2 + \angle 3 + \angle 4 = 360^\circ$. But $\angle 1 = \angle 3$ (vertical angles), $\angle 2 = \angle 4$ (vertical angles). So $2\angle 1 + 2\angle 2 = 360^\circ$, so $\angle 1 + \angle 2 = 180^\circ$. So $\angle 2 = 180^\circ - \angle 1 = 180^\circ - 50^\circ = 130^\circ$? Wait, that can't be. Wait, maybe I made a mistake. Wait, the top triangle: right angle, $40^\circ$, so $\angle 1 = 50^\circ$ (correct, because $90 + 40 + 50 = 180$). Then, the bottom triangle: right angle, so $\angle 5 = 40^\circ$, so $\angle 3 = 50^\circ$ (same as $\angle 1$). Then, at the intersection, $\angle 1 = 50^\circ$, $\angle 3 = 50^\circ$, so $\angle 2 = 180 - 50 = 130^\circ$? Wait, but that seems odd. Wait, maybe $\angle 2$ is equal to $\angle 1$? No, that doesn't make sense. Wait, maybe the two triangles are similar, so $\angle 1 = \angle 3 = 50^\circ$, and $\angle 2 = \angle 4 = 130^\circ$? Wait, but let's check problem 31: $\angle 3$. $\angle 3$ should be equal to $\angle 1$ (vertical angles), so $\angle 3 = 50^\circ$. Then problem 32: $\angle 4$ should be equal to $\angle 2$ (vertical angles), so $\angle 4 = 130^\circ$. Then problem 33: $\angle 5$. The bottom triangle is a right triangle, so $\angle 5 + 90^\circ + \angle 3 = 180^\circ$. $\angle 3 = 50^\circ$, so $\angle 5 = 180 - 90 - 50 = 40^\circ$. Then problem 34: $\angle 6$. $\angle 5$ and $\angle 6$ are supplementary (linear pair), so $\angle 6 = 180 - 40 = 140^\circ$? Wait, no, $\angle 5$ is $40^\circ$, and $\angle 6$ is adjacent to it, forming a linear pair, so $\angle 6 = 180 - 40 = 140^\circ$? Wait, no, the diagram shows $\angle 6$ as an external angle? Wait, maybe I need to re-examine.
Wait, let's start over with each problem:
Problem 29: $\angle 1$
Top triangle: right angle ($90^\circ$), angle of $40^\circ$, so $\angle 1 = 180 - 90 - 40 = 50^\circ$. So $\angle 1 = 50^\circ$.
Problem 30: $\angle 2$
$\angle 1$ and $\angle 2$ are supplementary (they form a linear pair, since they are adjacent and on a straight line). So $\angle 1 + \angle 2 = 180^\circ$. So $\angle 2 = 180 - 50 = 130^\circ$.
Problem 31: $\angle 3$
$\angle 3$ is vertical to $\angle 1$, so $\angle 3 = \angle 1 = 50^\circ$.
Problem 32: $\angle 4$
$\angle 4$ is vertical to $\angle 2$, so $\angle 4 = \angle 2 = 130^\circ$.
Problem 33: $\angle 5$
Bottom triangle: right angle ($90^\circ$), $\angle 3 = 50^\circ$, so $\angle 5 = 180 - 90 - 50 = 40^\circ$.
Problem 34: $\angle 6$
$\angle 5$ and $\angle 6$ are supplementary (linear pair), so $\angle 6 = 180 - 40 = 140^\circ$? Wait, no, $\angle 5$ is $40^\circ$, and $\angle 6$ is adjacent to it, forming a linear pair, so $\angle 6 = 180 - 40 = 140^\circ$? Wait, but the diagram shows $\angle 6$ as an external angle? Wait, maybe $\angle 5$ and $\angle 6$ are vertical to $\angle 8$ and $\angle 7$? Wait, problem 35: $\angle 7$ is a right angle? Wait, the diagram shows $\angle 7$ with a right angle symbol, so $\angle 7 = 90^\circ$. Then problem 36: $\angle 8$. $\angle 8$ and the $40^\circ$ angle are supplementary? Wait, $\angle 8$ is adjacent to the $40^\circ$ angle, forming a linear pair, so $\angle 8 = 180 - 40 = 140^\circ$? Wait, no, the top triangle: the $40^\circ$ angle and $\angle 8$ are supplementary, so $\angle 8 = 180 - 40 = 140^\circ$.
Let's summarize:
- $\angle 1 = 50^\circ$
- $\angle 2 = 130^\circ$
- $\angle 3 = 50^\circ$
- $\angle 4 = 130^\circ$
- $\angle 5 = 40^\circ$
- $\angle 6 = 140^\circ$
- $\angle 7 = 90^\circ$ (given by the right angle symbol)
- $\angle 8 = 140^\circ$ (supplementary to $40^\circ$)
Final Answers:
- $\boxed{50^\circ}$
- $\boxed{130^\circ}$
- $\boxed{50^\circ}$
- $\boxed{130^\circ}$
- $\boxed{40^\circ}$
- $\boxed{140^\circ}$
- $\boxed{90^\circ}$
- $\boxed{140^\circ}$