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GaryK [48]
3 years ago
15

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

Mathematics
2 answers:
Katyanochek1 [597]3 years ago
7 0

Answer:

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

Step-by-step 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 \frac{1}{2}m\stackrel{\frown}{DG}=m\angle DFG

We know that m∠DFG = 38°. So:

\displaystyle \frac{1}{2}m\stackrel{\frown}{DG}=38\Rightarrow m\stackrel{\frown}{DG}=76^\circ

∠DEG also intercepts Arc DG. Hence:

\displaystyle m\angle DEG=\frac{1}{2}m\stackrel{\frown}{DG}

We know that Arc DG measures 76°. Hence:

\displaystyle m\angle DEG =\frac{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 \frac{1}{2}m\stackrel{\frown}{AC}=m \angle ACB

Since m∠ACB = 70°:

\displaystyle \frac{1}{2}m\stackrel{\frown}{AC}=70\Rightarrow m\stackrel{\frown}{AC}=140^\circ

∠<em>x</em> 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 <em>n</em> 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 <em>y:</em>

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 \frac{1}{2}m\stackrel{\frown}{EDG}=m\angle EFG

Since m∠EFG = 115°:

\displaystyle \frac{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 = \frac{1}{2}m\stackrel{\frown}{EFG}

Since we know that Arc EFG measures 130°:

\displaystyle m\angle EDG = \frac{1}{2}(130)=65^\circ

scoundrel [369]3 years ago
4 0

Answer:

question 1:

38 degrees (explanation down below)

question 2:

y = 49 degrees!

question 3:

angle EDG = 65

Step-by-step 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

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Answer:

1. 18

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Step-by-step explanation:

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5 0
2 years ago
A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of te triangle is 16m to the power of 2, what are the base
Olenka [21]

Answer:

The base of triangle is  \frac{8}{\sqrt{3}} \ m and the height of triangle is  \frac{12}{\sqrt{3}} \ m.

Step-by-step explanation:

Given:

A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of the triangle is 16m to the power of 2.

Now, to find the base and height of the triangle.

The base of triangle = x\times\frac{1}{2} =\frac{x}{2} \ m.

The height of triangle = x\times \frac{3}{4} =\frac{3x}{4}\ m.

The area of triangle = 16\ m^2.

Now, we put the formula of area to solve:

Area=\frac{1}{2} \times base\times height

16=\frac{1}{2} \times \frac{x}{2} \times \frac{3x}{4}

16=\frac{3x^2}{16}

<em>Multiplying both sides by 16 we get:</em>

<em />256=3x^2<em />

<em>Dividing both sides by 3 we get:</em>

<em />\frac{256}{3} =x^2<em />

<em>Using square root on both sides we get:</em>

\frac{16}{\sqrt{3}}=x

x=\frac{16}{\sqrt{3}}

Now, by substituting the value of x to get the base and height:

Base=\frac{x}{2}\\\\Base=\frac{\frac{16}{\sqrt{3}}}{2} \\\\Base=\frac{8}{\sqrt{3}} \ m.

<em>So, the base of triangle = </em>\frac{8}{\sqrt{3}} \ m.<em />

Height=x\times\frac{3}{4} \\\\Height=\frac{16}{\sqrt{3}}\times \frac{3}{4} \\\\Height=\frac{12}{\sqrt{3}} \ m.

<em>Thus, the height of triangle =  </em>\frac{12}{\sqrt{3}} \ m.<em />

Therefore, the base of triangle is  \frac{8}{\sqrt{3}} \ m and the height of triangle is  \frac{12}{\sqrt{3}} \ m.

7 0
3 years ago
Write the equation of a line that is perpendicular to y=3x-2 and that passes through the point (-9,5)
damaskus [11]
Slope of perpendicular is -1/3.
Then equation is
y-5=(-1/3)(x+9)
8 0
2 years ago
I PROMISE BRANLIEST TO THE FIRST ANSWER SUBMITTED!
Vedmedyk [2.9K]

Answer:

36

Step-by-step explanation:

because square of term on LHS = square of term on RHS

if 5 - 3y = 6

then (5 - 3y)² = 36

8 0
2 years ago
Two events E1 and E2 are called independent if p(E1 â© E2) = p(E1)p(E2). For each of the following pairs of events, which are su
posledela

Answer:  a) Independent

b) Independent

c) Dependent

Step-by-step explanation:

Since, If a coin is tossed three times,

Then, total number of outcomes, n(S) = 8

a)  E_1 : tails comes up with the coin is tossed the first time;

E_1 = { TTT, THH, THT, TTH }

E_2 :  heads comes up when the coin is tossed the second time.

E_2 = { THT, HHH, THH, HHT }

Thus, n(E_1)=4

⇒  P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}

Similarly,  P(E_2)=\frac{1}{2}

⇒  P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}

Since,   E_1\cap E_2 = { THH, THT }

n(E_1\cap E_2) = 2

⇒ P(E_1\cap E_2) = \frac{n(E_1\cap E_2)}{n(S)}= \frac{2}{8}=\frac{1}{4}

Thus,  P(E_1\cap E_2)=P(E_1)\timesP(E_2)

Therefore,  E_1 and E_2 are independent events.

B)  E_1 :  the first coin comes up tails

E_1 = { TTT, THH, THT, TTH }

E_2 :   two, and not three, heads come up in a row

E_2 = { HHT, THH }

Thus, n(E_1)=4

⇒  P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}

Similarly,  P(E_2)=\frac{1}{4}

⇒  P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}

Since,   E_1\cap E_2 = { THH }

n(E_1\cap E_2) = 1

⇒ P(E_1\cap E_2) = \frac{n(E_1\cap E_2)}{n(S)}= \frac{1}{8}

Thus,  P(E_1\cap E_2)=P(E_1)\timesP(E_2)

Therefore,  E_1 and E_2 are independent events.

C)  E_1 :  the second coin comes up tails;

E_1 = { HTH, HTT, TTT, TTH }

E_2 :   two, and not three, heads come up in a row

E_2 = { HHT, THH }

Thus, n(E_1)=4

⇒  P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}

Similarly,  P(E_2)=\frac{1}{4}

⇒  P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}

Since,   E_1\cap E_2 = \phi

n(E_1\cap E_2) = 0

⇒ P(E_1\cap E_2) = 0

Thus,  P(E_1\cap E_2)\neq P(E_1)\timesP(E_2)

Therefore,  E_1 and E_2 are dependent events.


3 0
3 years ago
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