13. if $m\angle fgh = (6x + 21)^\circ$ and $m\overarc{fjh} = (17x - 28)^\circ$, find $m\overarc{fjh}$.

14. if $m\angle stu = (5x - 16)^\circ$ and $m\overarc{su} = (12x - 50)^\circ$ , find $m\angle stu$.

15. if $m\angle abd = (6x + 26)^\circ$ and $m\angle acd = (13x - 9)^\circ$ , find $m\overarc{ad}$.

16. if $m\angle kjl = (3x + 2)^\circ$ and $m\angle klj = (7x - 32)^\circ$ , find $m\overarc{kl}$.

Question

13. if $m\angle fgh = (6x + 21)^\circ$ and $m\overarc{fjh} = (17x - 28)^\circ$, find $m\overarc{fjh}$.

14. if $m\angle stu = (5x - 16)^\circ$ and $m\overarc{su} = (12x - 50)^\circ$ , find $m\angle stu$.

15. if $m\angle abd = (6x + 26)^\circ$ and $m\angle acd = (13x - 9)^\circ$ , find $m\overarc{ad}$.

16. if $m\angle kjl = (3x + 2)^\circ$ and $m\angle klj = (7x - 32)^\circ$ , find $m\overarc{kl}$.

Accepted Answer

### Problem 13
# Explanation:
## Step1: Recall Inscribed Angle Theorem
The measure of an inscribed angle is half the measure of its intercepted arc. So, $ m\angle FGH=\frac{1}{2}m\widehat{FJH} $.
Substitute the given expressions: $ 6x + 21=\frac{1}{2}(17x - 28) $.

## Step2: Solve for $ x $
Multiply both sides by 2: $ 2(6x + 21)=17x - 28 $
$ 12x + 42 = 17x - 28 $
Subtract $ 12x $ from both sides: $ 42 = 5x - 28 $
Add 28 to both sides: $ 70 = 5x $
Divide by 5: $ x = 14 $.

## Step3: Find $ m\widehat{FJH} $
Substitute $ x = 14 $ into $ 17x - 28 $:
$ 17(14)-28 = 238 - 28 = 210 $.

# Answer: $ 210^\circ $

### Problem 14
# Explanation:
## Step1: Apply Inscribed Angle Theorem
$ m\angle STU=\frac{1}{2}m\widehat{SU} $, so $ 5x - 16=\frac{1}{2}(12x - 50) $.

## Step2: Solve for $ x $
Multiply both sides by 2: $ 2(5x - 16)=12x - 50 $
$ 10x - 32 = 12x - 50 $
Subtract $ 10x $: $ -32 = 2x - 50 $
Add 50: $ 18 = 2x $
Divide by 2: $ x = 9 $.

## Step3: Find $ m\angle STU $
Substitute $ x = 9 $ into $ 5x - 16 $:
$ 5(9)-16 = 45 - 16 = 29 $.

# Answer: $ 29^\circ $

### Problem 15
# Explanation:
## Step1: Recognize Inscribed Angles Intercepting Same Arc
$ \angle ABD $ and $ \angle ACD $ intercept $ \widehat{AD} $, so they are equal. Thus, $ 6x + 26 = 13x - 9 $.

## Step2: Solve for $ x $
Subtract $ 6x $: $ 26 = 7x - 9 $
Add 9: $ 35 = 7x $
Divide by 7: $ x = 5 $.

## Step3: Find $ m\widehat{AD} $
Let $ m\widehat{AD}=y $. The inscribed angle $ \angle ABD=\frac{1}{2}y $, so $ 6(5)+26=\frac{1}{2}y $
$ 30 + 26=\frac{1}{2}y $
$ 56=\frac{1}{2}y $
Multiply by 2: $ y = 112 $.

# Answer: $ 112^\circ $

### Problem 16
# Explanation:
## Step1: Analyze Triangle $ KJL $ and Inscribed Angle
In $ \triangle KJL $, $ \angle KJL $ and $ \angle KLJ $ are angles, and $ \angle KJL $ intercepts $ \widehat{KL} $, $ \angle KLJ $ intercepts $ \widehat{KJ} $. But since $ JL $ is a diameter (center $ M $), $ \angle JK L = 90^\circ $ (Thales' theorem). Wait, actually, $ \angle KJL $ and $ \angle KLJ $ are angles in triangle, but also, the inscribed angle over $ \widehat{KL} $ is related. Wait, correct approach: Inscribed angle over $ \widehat{KL} $ is $ \angle KJL $, so $ m\angle KJL=\frac{1}{2}m\widehat{KL} $? No, wait, $ \angle KJL $ intercepts $ \widehat{KL} $, $ \angle KLJ $ intercepts $ \widehat{KJ} $. But since $ JL $ is diameter, $ m\widehat{KJ}+m\widehat{KL}=180^\circ $. Also, in triangle $ KJL $, angles sum to $ 180^\circ $, so $ \angle KJL+\angle KLJ+\angle JK L = 180^\circ $, and $ \angle JK L = 90^\circ $ (diameter subtends right angle). So $ (3x + 2)+(7x - 32)+90 = 180 $.

## Step2: Solve for $ x $
Combine like terms: $ 10x + 60 = 180 $
Subtract 60: $ 10x = 120 $
Divide by 10: $ x = 12 $.

## Step3: Find $ m\widehat{KL} $
$ \angle KJL $ intercepts $ \widehat{KL} $, so $ m\angle KJL=\frac{1}{2}m\widehat{KL} $? Wait, no, $ \angle KJL $ is an inscribed angle intercepting $ \widehat{KL} $, so $ m\angle KJL=\frac{1}{2}m\widehat{KL} $. Wait, earlier mistake: Thales' theorem says $ \angle JK L = 90^\circ $, so triangle $ KJL $ is right-angled at $ K $. So $ \angle KJL+\angle KLJ = 90^\circ $. So $ (3x + 2)+(7x - 32)=90 $.
That's the correct approach! So:
$ 10x - 30 = 90 $
$ 10x = 120 $
$ x = 12 $.
Now, $ \angle KJL = 3(12)+2 = 38^\circ $, which is an inscribed angle over $ \widehat{KL} $, so $ m\widehat{KL}=2\times38 = 76^\circ $. Wait, or $ \angle KLJ = 7(12)-32 = 52^\circ $, inscribed over $ \widehat{KJ} $, so $ m\widehat{KJ}=104^\circ $, and since $ JL $ is diameter, $ m\widehat{KL}+m\widehat{KJ}=180^\circ $, $ 76 + 104 = 180 $, correct. So using $ \angle KJL $: $ m\widehat{KL}=2\times m\angle KJL = 2\times(3x + 2) $. Substitute $ x = 12 $: $ 2\times(38)=76 $.

# Answer: $ 76^\circ $

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QUESTION IMAGE

Question

Explanation:

Problem 13

Step1: Recall Inscribed Angle Theorem

Step2: Solve for \( x \)

Step3: Find \( m\widehat{FJH} \)

Step1: Apply Inscribed Angle Theorem

Step2: Solve for \( x \)

Step3: Find \( m\angle STU \)

Step1: Recognize Inscribed Angles Intercepting Same Arc

Step2: Solve for \( x \)

Step3: Find \( m\widehat{AD} \)

Answer:

Problem 14