Question was provided via image upload.

Question

Accepted Answer

### Problem 4
# Explanation:
To determine the measure of $ w^\circ $, we use the fact that the sum of the interior angles of a triangle is $ 180^\circ $, and a linear pair of angles also sums to $ 180^\circ $. The two given interior angles are $ 42^\circ $ and $ 39^\circ $. First, find the third interior angle: $ 180 - (42 + 39) = 99^\circ $. Then, $ w $ and this $ 99^\circ $ angle form a linear pair, so $ w = 180 - 99 = 81^\circ $? Wait, no—wait, looking at the options:

Wait, option a: $ w = 99^\circ $, because $ 42 + 39 = 81 $ and $ 180 - 81 = 99 $. Wait, maybe I misread the diagram. If $ w $ is an exterior angle? Wait, no, the diagram shows a triangle with angles $ 42^\circ $, $ 39^\circ $, and the third angle, and $ w $ is adjacent to the third angle. Wait, the sum of interior angles is $ 180 $. So $ 42 + 39 + \text{third angle} = 180 $, so third angle is $ 99 $. Then $ w $ and the third angle are supplementary (linear pair), so $ w = 180 - 99 = 81 $? But option a says $ w = 99 $. Wait, maybe the diagram is different. Wait, the options:

a. $ w = 99^\circ $, because $ 42 + 39 = 81 $ and $ 180 - 81 = 99 $.

b. $ w = 261^\circ $, which is wrong (sum of angles can't exceed 180 for a triangle, and linear pair is 180).

c. $ w = 81^\circ $, because $ 180 - (42 + 39) = 99 $ and $ 180 - 99 = 81 $.

Wait, maybe the diagram has $ w $ as the exterior angle? Wait, no—wait, if $ w $ is an exterior angle, then it's equal to the sum of the two non-adjacent interior angles. So $ w = 42 + 39 = 81 $? But option a says 99. Wait, maybe the diagram is a triangle with angles 42, 39, and the third angle, and $ w $ is the third angle? Wait, that would be $ 42 + 39 + w = 180 $, so $ w = 99 $. Ah, that must be it. So the diagram shows $ w $ as one of the interior angles? Wait, the user's diagram: "a triangle with angles 42°, 39°, and w°"? No, the diagram has 42°, 39°, and $ w $ is adjacent. Wait, the options:

a. $ w = 99^\circ $, because $ 42 + 39 = 81 $ and $ 180 - 81 = 99 $. So that's correct: sum of 42 and 39 is 81, so the third angle (w) is 180 - 81 = 99. So option a is correct. Wait, but the user marked option c. Wait, maybe I misread. Let's re-express:

For a triangle, sum of interior angles is $ 180^\circ $. So if two angles are 42 and 39, the third is $ 180 - 42 - 39 = 99 $. So if $ w $ is that third angle, then $ w = 99 $, which is option a. So the correct answer is a.

# Answer:
a. $ w = 99^\circ $, because $ 42 + 39 = 81 $ and $ 180 - 81 = 99 $

### Problem 5
# Explanation:
We have triangle $ DCE $ with $ \angle D = 2x $, $ \angle C = 70^\circ $, and $ \angle DEC = 4x + 10 $? Wait, no—wait, the diagram shows $ \angle REC = 4x + 10 $, and $ \angle D = 2x $, $ \angle C = 70^\circ $, and $ \angle DEC $ and $ \angle REC $ are supplementary (linear pair), so $ \angle DEC + \angle REC = 180^\circ $. Also, in triangle $ DCE $, $ \angle D + \angle C + \angle DEC = 180^\circ $, so $ 2x + 70 + \angle DEC = 180 $, so $ \angle DEC = 110 - 2x $. But $ \angle DEC + (4x + 10) = 180 $, so $ (110 - 2x) + (4x + 10) = 180 $. Simplify: $ 120 + 2x = 180 $, so $ 2x = 60 $, $ x = 30 $. Then $ \angle REC = 4(30) + 10 = 130^\circ $? Wait, but the number bank has 130. Wait, maybe I messed up. Wait, the equation: $ 2x + 70 = 4x + 10 $ (exterior angle theorem: exterior angle equals sum of two remote interior angles). Yes! The exterior angle theorem: $ \angle REC $ is an exterior angle to triangle $ DCE $, so it equals the sum of the two non-adjacent interior angles, $ \angle D $ and $ \angle C $. So $ 4x + 10 = 2x + 70 $. Solve: $ 4x - 2x = 70 - 10 $, $ 2x = 60 $, $ x = 30 $. Then $ \angle REC = 4(30) + 10 = 130^\circ $.

## Step 1: Apply Exterior Angle Theorem
$ 4x + 10 = 2x + 70 $

## Step 2: Solve for $ x $
$ 4x - 2x = 70 - 10 $
$ 2x = 60 $
$ x = 30 $

## Step 3: Find $ \angle REC $
$ \angle REC = 4(30) + 10 = 130^\circ $

# Answer:
$ x = 30 $, $ \angle REC = 130^\circ $

### Problem 6
# Explanation:
In triangle $ TSU $, $ \angle T = 81^\circ $, $ \angle S = 12x $, and $ \angle TUV = 24x + 9 $ is an exterior angle. By exterior angle theorem, $ \angle TUV = \angle T + \angle S $, so $ 24x + 9 = 81 + 12x $. Solve for $ x $:

## Step 1: Set up equation
$ 24x + 9 = 81 + 12x $

## Step 2: Solve for $ x $
$ 24x - 12x = 81 - 9 $
$ 12x = 72 $
$ x = 6 $

## Step 3: Find $ \angle TSU $ (which is $ 12x $)
$ 12(6) = 72^\circ $

# Answer:
$ x = 6 $, $ \angle TSU = 72^\circ $

### Problem 7
# Explanation:
In triangle $ JKL $, $ \angle J = 39^\circ $, $ \angle L = 2x + 1 $, and $ \angle KAL = 3x + 13 $ is an exterior angle. By exterior angle theorem, $ 3x + 13 = 39 + (2x + 1) $. Solve:

## Step 1: Set up equation
$ 3x + 13 = 39 + 2x + 1 $

## Step 2: Simplify
$ 3x + 13 = 2x + 40 $

## Step 3: Solve for $ x $
$ 3x - 2x = 40 - 13 $
$ x = 27 $

## Step 4: Find $ \angle JLK $ (which is $ 2x + 1 $)
$ 2(27) + 1 = 55^\circ $

# Answer:
$ x = 27 $, $ \angle JLK = 55^\circ $

### Problem 8
# Explanation:
In triangle $ FHG $, $ \angle H = 95^\circ $ (wait, no—$ \angle IHG = 95^\circ $ is an exterior angle to triangle $ FHG $, so $ \angle IHG = \angle F + \angle G $. $ \angle F = 4x - 8 $, $ \angle G = 3x - 16 $. So:

## Step 1: Set up equation (exterior angle theorem)
$ 95 = (4x - 8) + (3x - 16) $

## Step 2: Simplify
$ 95 = 7x - 24 $

## Step 3: Solve for $ x $
$ 7x = 95 + 24 = 119 $
$ x = 17 $

## Step 4: Find $ \angle FGH $ (which is $ 3x - 16 $)
$ 3(17) - 16 = 51 - 16 = 35^\circ $

# Answer:
$ x = 17 $, $ \angle FGH = 35^\circ $

### Problem 9
# Explanation:
In triangle $ XYZ $, $ \angle X = 5x + 33 $, $ \angle Y = 5x + 2 $, and $ \angle ZBZ = 14x + 3 $ is an exterior angle. By exterior angle theorem, $ 14x + 3 = (5x + 33) + (5x + 2) $. Solve:

## Step 1: Set up equation
$ 14x + 3 = 10x + 35 $

## Step 2: Solve for $ x $
$ 14x - 10x = 35 - 3 $
$ 4x = 32 $
$ x = 8 $

## Step 3: Find $ \angle BZY $ (which is $ 14x + 3 $)
$ 14(8) + 3 = 115^\circ $

# Answer:
$ x = 8 $, $ \angle BZY = 115^\circ $

### Problem 10
# Explanation:
In triangle $ QPR $, the angles are $ (-17x - 1)^\circ $, $ (-10x + 17)^\circ $, and $ (-32x + 1)^\circ $ (since $ \angle ERP $ is vertical to the angle at $ R $, so they are equal? Wait, sum of interior angles is $ 180^\circ $. So:

$ (-17x - 1) + (-10x + 17) + (-32x + 1) = 180 $

## Step 1: Combine like terms
$ (-17x - 10x - 32x) + (-1 + 17 + 1) = 180 $
$ -59x + 17 = 180 $

## Step 2: Solve for $ x $
$ -59x = 163 $
Wait, that can't be. Wait, maybe $ \angle ERP $ is an exterior angle. Wait, the angles are $ \angle Q = -17x - 1 $, $ \angle P = -10x + 17 $, and $ \angle R = -32x + 1 $ (but $ \angle ERP $ is adjacent to $ \angle R $, so they are supplementary? No, sum of interior angles:

Wait, let's check the number bank. Maybe I made a mistake. Let's try again:

Sum of angles: $ (-17x - 1) + (-10x + 17) + (-32x + 1) = 180 $
Combine: $ -59x + 17 = 180 $ → $ -59x = 163 $ → $ x = -163/59 $, which is not in the number bank. So maybe exterior angle. Let's see: $ \angle ERP = \angle Q + \angle P $, so $ -32x + 1 = (-17x - 1) + (-10x + 17) $
Simplify: $ -32x + 1 = -27x + 16 $
$ -32x + 27x = 16 - 1 $
$ -5x = 15 $
$ x = -3 $

Then $ \angle ERP = -32(-3) + 1 = 96 + 1 = 97^\circ $, which is in the number bank. Yes! So:

## Step 1: Set up exterior angle theorem
$ -32x + 1 = (-17x - 1) + (-10x + 17) $

## Step 2: Simplify right side
$ -27x + 16 $

## Step 3: Solve for $ x $
$ -32x + 1 = -27x + 16 $
$ -32x + 27x = 16 - 1 $
$ -5x = 15 $
$ x = -3 $

## Step 4: Find $ \angle ERP $
$ -32(-3) + 1 = 97^\circ $

# Answer:
$ x = -3 $, $ \angle ERP = 97^\circ $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 4

Step 1: Apply Exterior Angle Theorem

Step 2: Solve for \( x \)

Step 3: Find \( \angle REC \)

Step 1: Set up equation

Step 2: Solve for \( x \)

Step 3: Find \( \angle TSU \) (which is \( 12x \))

Answer:

Problem 5