Question

All 3 questions 3 pictures for find the missing angles, please with reasoning why? 100 points

Answer 1

question 1:

38 degrees (explanation down below)

question 2:

y = 49 degrees!

question 3:

angle EDG = 65

Explanation:

question 1: i'd say DFG is the same as DEG because the arc is the same angle, so angle DEG is 38 degrees because it is half of 76 degrees or the arc angle

question 2 isn't any harder: the arc angle for 70 is 140 degrees, so 140 degrees is equal to x. so angle O would be 220 degrees and because the triangle became a square, 360-220-70-21 equals y!

question 3: arc angles teaches you something that will make your question SO easy, arc angle GE is equal to 230 degrees because of 115 degrees doubled, so 360/230 equals 130 which divided by 2 equals 65

Answer 2

Problem 1)

`m\angle DEG = 38^\circ`

Problem 2)

`m\angle x =140^\circ \text{ and } m\angle y = 49^\circ`

Problem 3)

`m\angle EDG = 65^\circ`

Explanation:

Problem 1: ∠DEG

From the diagram, we know that ∠DFG intercepts Arc DG.

Inscribed angles have half the measure of its intercepted arc. Therefore:

`\displaystyle (1)/(2)m\stackrel{\frown}{DG}=m\angle DFG`

We know that m∠DFG = 38°. So:

`\displaystyle (1)/(2)m\stackrel{\frown}{DG}=38\Rightarrow m\stackrel{\frown}{DG}=76^\circ`

∠DEG also intercepts Arc DG. Hence:

`\displaystyle m\angle DEG=(1)/(2)m\stackrel{\frown}{DG}`

We know that Arc DG measures 76°. Hence:

`\displaystyle m\angle DEG =(1)/(2)\left(76\right)=38^\circ`

Alternate Explanation:

Since ∠DEG and ∠DFG intercept the same arc, ∠DEG ≅ DFG. So, m∠DEG = m∠DFG = 38°.

Problem 2: Circle with Centre O

(Let the bottom left corner be A, upper be B, and right be C.)

∠ACB intercepts Arc AC.

Inscribed angles have half the measure of its intercepted arc. Therefore:

`\displaystyle (1)/(2)m\stackrel{\frown}{AC}=m \angle ACB`

Since m∠ACB = 70°:

`\displaystyle (1)/(2)m\stackrel{\frown}{AC}=70\Rightarrow m\stackrel{\frown}{AC}=140^\circ`

∠x is a central angle and also intercepts Arc AC.

The measure of a central angle is equal to its intercepted arc. Thus:

`\displaystyle m\angle x =m\stackrel{\frown}{AC}=140^\circ`

The sum of the interior angles of a polygon is given by the formula:

`(n-2)\cdot 180^\circ`

Where n is the number of sides.

Since the inscribed figure is a four-sided polygon, its interior angles must total:

`(4-2)\cdot 180^\circ =360^\circ`

Therefore:

`21+70+y+(360-x)=360`

Substitute and solve for y:

`91+y+(360-140)=360\Rightarrow y +311=360\Rightarrow m\angle y = 49^\circ`

Problem 3: ∠EDG

∠EFG intercepts Arc EDG.

Inscribed angles have half the measure of its intercepted arc. Therefore:

`\displaystyle (1)/(2)m\stackrel{\frown}{EDG}=m\angle EFG`

Since m∠EFG = 115°:

`\displaystyle (1)/(2)m\stackrel{\frown}{EDG} = 115\Rightarrow m\stackrel{\frown}{EDG} = 230^\circ`

A full circle measures 360°. Hence:

`m\stackrel{\frown}{EDG}+m\stackrel{\frown}{EFG}=360^\circ`

Since we know that Arc EDG measures 230°:

`230^\circ + m\stackrel{\frown}{EFG}=360`

Solve for Arc EFG:

`m\stackrel{\frown}{EFG}=130^\circ`

∠EDG intercepts Arc EFG.

Inscribed angles have half the measure of its intercepted arc. Therefore:

`\displaystyle m\angle EDG = (1)/(2)m\stackrel{\frown}{EFG}`

Since we know that Arc EFG measures 130°:

`\displaystyle m\angle EDG = (1)/(2)(130)=65^\circ`