find the measure of the angle indicated.
11) find ( mangle wst ).
12) find ( mangle r ).
13) find ( mangle mfg ).
14) find ( mangle efd ).
find the measure of each angle indicated.
15)
16)
solve for ( x ).
17)
18)
find the measure of angle ( a ).
19)
20)
find the measure of each angle indicated.
21)
22)

Question

Accepted Answer

### Problem 15: Find the measure of the indicated angle (triangle with 80° and 35°)
# Explanation:
## Step 1: Recall Triangle Angle Sum Theorem
The sum of the interior angles of a triangle is $ 180^\circ $. Let the unknown angle be $ y $. So, $ 35^\circ + 80^\circ + y = 180^\circ $.
## Step 2: Solve for $ y $
First, add $ 35^\circ $ and $ 80^\circ $: $ 35 + 80 = 115 $. Then, $ y = 180 - 115 = 65^\circ $.
# Answer: $ 65^\circ $

### Problem 16: Find the measure of the indicated angle (triangle with two 70° angles)
# Explanation:
## Step 1: Recall Triangle Angle Sum Theorem
Let the unknown angle be $ z $. So, $ 70^\circ + 70^\circ + z = 180^\circ $.
## Step 2: Solve for $ z $
Add $ 70^\circ $ and $ 70^\circ $: $ 70 + 70 = 140 $. Then, $ z = 180 - 140 = 40^\circ $.
# Answer: $ 40^\circ $

### Problem 17: Solve for $ x $ (triangle with angles $ x + 70 $, $ 76^\circ $, $ 41^\circ $)
# Explanation:
## Step 1: Recall Triangle Angle Sum Theorem
$ (x + 70) + 76 + 41 = 180 $.
## Step 2: Simplify and solve for $ x $
Combine like terms: $ x + 70 + 76 + 41 = x + 187 $. So, $ x + 187 = 180 $. Subtract 187 from both sides: $ x = 180 - 187 = -7 $. Wait, that seems odd. Wait, maybe a typo? Wait, no, let's check again. Wait, $ 76 + 41 = 117 $, so $ x + 70 + 117 = 180 $ → $ x + 187 = 180 $ → $ x = -7 $. Hmm, negative angle? Maybe the problem has a typo, but following the math, $ x = -7 $.
# Answer: $ x = -7 $

### Problem 18: Solve for $ x $ (right triangle with angles $ 25^\circ $, $ 90^\circ $, $ x + 69 $)
# Explanation:
## Step 1: Recall Triangle Angle Sum Theorem (right triangle, so one angle is $ 90^\circ $)
$ 25 + 90 + (x + 69) = 180 $.
## Step 2: Solve for $ x $
Add $ 25 + 90 = 115 $, then $ 115 + x + 69 = x + 184 $. So, $ x + 184 = 180 $ → $ x = 180 - 184 = -4 $. Again, negative. Maybe a typo, but math gives $ x = -4 $.
# Answer: $ x = -4 $

### Problem 19: Find $ m\angle A $ (triangle with angles $ 57 + x $, $ x + 87 $, $ 50^\circ $)
# Explanation:
## Step 1: Recall Triangle Angle Sum Theorem
$ (57 + x) + (x + 87) + 50 = 180 $.
## Step 2: Simplify and solve for $ x $
Combine like terms: $ 57 + x + x + 87 + 50 = 2x + 194 $. So, $ 2x + 194 = 180 $ → $ 2x = 180 - 194 = -14 $ → $ x = -7 $.
## Step 3: Find $ m\angle A $ (which is $ 57 + x $)
Substitute $ x = -7 $: $ 57 + (-7) = 50^\circ $. Wait, but let's check the other angle: $ x + 87 = -7 + 87 = 80^\circ $. Then $ 50 + 80 + 50 = 180 $? Wait, $ 50 + 80 + 50 = 180 $? No, $ 50 + 80 + 50 = 180 $? 50 + 80 is 130, +50 is 180. Oh, right! So $ m\angle A = 50^\circ $.
# Answer: $ m\angle A = 50^\circ $

### Problem 20: Find $ m\angle A $ (triangle with angles $ 80^\circ $, $ 9x + 4 $, $ 16x - 4 $)
# Explanation:
## Step 1: Recall Triangle Angle Sum Theorem
$ (9x + 4) + (16x - 4) + 80 = 180 $.
## Step 2: Simplify and solve for $ x $
Combine like terms: $ 9x + 16x + 4 - 4 + 80 = 25x + 80 $. So, $ 25x + 80 = 180 $ → $ 25x = 100 $ → $ x = 4 $.
## Step 3: Find $ m\angle A $ (which is $ 9x + 4 $)
Substitute $ x = 4 $: $ 9(4) + 4 = 36 + 4 = 40^\circ $.
# Answer: $ m\angle A = 40^\circ $

### Problem 21: Find the indicated angle (two triangles, first has $ 30^\circ $, $ 100^\circ $, $ 70^\circ $; second has $ 60^\circ $, $ 70^\circ $,? )
# Explanation:
## Step 1: Recall Triangle Angle Sum Theorem for the second triangle
The angle adjacent to $ 70^\circ $ is vertical? Wait, no, the first triangle: $ 30 + 100 + 50 = 180 $? Wait, no, $ 30 + 100 + 50 = 180 $, but the angle at the vertex is $ 70^\circ $, so the adjacent angle (linear pair) would be $ 180 - 70 = 110^\circ $? Wait, no, the diagram: two triangles sharing a common angle? Wait, the first triangle has angles $ 30^\circ $, $ 100^\circ $, so the third angle is $ 180 - 30 - 100 = 50^\circ $. Then the angle between the two triangles is $ 70^\circ $, so the angle in the second triangle adjacent to $ 70^\circ $ is $ 180 - 50 - 70 = 60^\circ $? Wait, no, the second triangle has angles $ 60^\circ $, $ 70^\circ $? Wait, no, let's re-express. The first triangle: angles $ 30^\circ $, $ 100^\circ $, so the third angle (at the base) is $ 50^\circ $. Then the angle between the two triangles is $ 70^\circ $, so the angle in the second triangle is $ 180 - 50 - 70 = 60^\circ $? Wait, no, the second triangle has angles $ 60^\circ $, $ 70^\circ $, so the third angle is $ 180 - 60 - 70 = 50^\circ $? Wait, maybe the diagram is two triangles with a common vertex angle. Wait, the first triangle: $ 30^\circ $, $ 100^\circ $, so the angle at the vertex (between the two triangles) is $ 180 - 30 - 100 = 50^\circ $? No, the given angle is $ 70^\circ $. Wait, maybe the sum of angles around a point: the angles around the common vertex are $ 30 + 100 + 70 + 60 +? = 360 $? No, it's two triangles. Wait, the first triangle: angles $ 30^\circ $, $ 100^\circ $, so the third angle is $ 50^\circ $. Then the angle adjacent to $ 50^\circ $ is $ 70^\circ $ (linear pair), so $ 180 - 50 - 70 = 60^\circ $. Then the second triangle has angles $ 60^\circ $, $ 60^\circ $, so the third angle is $ 180 - 60 - 60 = 60^\circ $? Wait, no, the second triangle has one angle $ 60^\circ $, and the angle adjacent to $ 70^\circ $ is $ 180 - 70 = 110^\circ $? I think I'm overcomplicating. Let's use triangle angle sum. The second triangle: angles $ 60^\circ $, and the angle opposite? Wait, the diagram: first triangle has $ 30^\circ $, $ 100^\circ $, so the angle at the base is $ 50^\circ $. Then the angle between the two triangles is $ 70^\circ $, so the angle in the second triangle is $ 180 - 60 - (180 - 30 - 100 - 70) $? No, better: the sum of angles in a triangle is $ 180 $. The second triangle has angles $ 60^\circ $, and the angle adjacent to the $ 70^\circ $ angle. Wait, the first triangle: $ 30 + 100 + 50 = 180 $. The angle between the two triangles is $ 70^\circ $, so the angle in the second triangle is $ 180 - 60 - (180 - 50 - 70) = 180 - 60 - 60 = 60^\circ $? No, that's not right. Wait, maybe the angle is $ 50^\circ $? No, let's do it properly. The first triangle: angles $ 30^\circ $, $ 100^\circ $, so the third angle (let's call it $ \angle 1 $) is $ 180 - 30 - 100 = 50^\circ $. Then, $ \angle 1 $, $ 70^\circ $, and the angle in the second triangle (let's call it $ \angle 2 $) are linear? No, they form a triangle. Wait, the second triangle has angles $ 60^\circ $, $ \angle 2 $, and the angle adjacent to $ 70^\circ $. Wait, maybe the diagram is two triangles with a common angle. The first triangle: $ 30^\circ $, $ 100^\circ $, $ 50^\circ $. The second triangle: $ 60^\circ $, $ 70^\circ $, so the third angle is $ 180 - 60 - 70 = 50^\circ $. Yes, that makes sense. So the indicated angle is $ 50^\circ $.
# Answer: $ 50^\circ $

### Problem 22: Find the indicated angle (two triangles with angles $ 30^\circ $, $ 40^\circ $, $ 25^\circ $)
# Explanation:
## Step 1: Use vertical angles and triangle angle sum
The vertical angles are equal. The first triangle has angles $ 30^\circ $, $ 40^\circ $, so the third angle is $ 180 - 30 - 40 = 110^\circ $ (vertical angle). Then the second triangle has angles $ 25^\circ $, $ 110^\circ $, so the indicated angle is $ 180 - 25 - 110 = 45^\circ $.
# Answer: $ 45^\circ $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 15: Find the measure of the indicated angle (triangle with 80° and 35°)

Step 1: Recall Triangle Angle Sum Theorem

Step 2: Solve for \( y \)

Step 1: Recall Triangle Angle Sum Theorem

Step 2: Solve for \( z \)

Step 1: Recall Triangle Angle Sum Theorem

Step 2: Simplify and solve for \( x \)

Answer:

Problem 16: Find the measure of the indicated angle (triangle with two 70° angles)