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In the figure below, \overline{AB}
AB
start overline, A, B, end overline is a diameter of circle PPP.
What is the arc measure of \stackrel{\LARGE{\frown}}{AD}
AD
⌢
A, D, start superscript, \frown, end superscript in degrees?

Answer :

TaeKwonDoIsDead

Answer:

Arc AD = 43°

Step-by-step explanation:

Image is attached.

When an angle subtends from an Arc to the center, the measure of the Arc and the Angle are same. Now, looking at the figure,

Arc AB = 180 (since this is an Arc subtending the diameter)

Arc DC = 90 degrees since the central angle is 90

So,

Arc AD + Arc CB = 180 - 90 = 90

Arc AD and Arc CB have the same measures as the central angle it created (terms in x). So we can write:

Arc AD + Arc CB = 90

7x + 1 + 9x - 7 = 90

We can solve for x:

[tex]7x + 1 + 9x - 7 = 90\\16x-6=90\\16x=96\\x=6[/tex]

The central angle AD is:

AD = 7x + 1 = 7(6) + 1 = 43

Also, Arc AD is same, thus:

Arc AD = 43°

${teks-lihat-gambar} TaeKwonDoIsDead

Answer:

m∠APD+m∠DPC+m∠CPD

(7x+1)+90+(9x−7)

16x+84

16x

x

​

 

=180

=180

=180

=96

=6

​

Hint #33 / 5

We can use the value of xxx to evaluate the measure of \angle APD∠APDangle, A, P, D. Let's substitute in our value for xxx.

\begin{aligned} m\angle APD&= (7x+1)^\circ \\\\ &=(7(6)+1)^\circ \\\\ &=43^\circ \end{aligned}

m∠APD

​

 

=(7x+1)

∘

=(7(6)+1)

∘

=43

∘

​

Hint #44 / 5

There are 360^\circ360

∘

360, degrees in a circle, so the measure of any major arc is equal to 360^\circ360

∘

360, degrees minus the measure of the corresponding minor arc.

m\stackrel{\large{\frown}}{ACD} \,= 360^\circ-43^\circm

ACD

⌢

=360

∘

−43

∘

m, A, C, D, start superscript, \frown, end superscript, equals, 360, degrees, minus, 43, degrees

Hint #55 / 5

The arc measure of \stackrel{\large{\frown}}{ACD}

ACD

⌢

A, C, D, start superscript, \frown, end superscript is 317^\circ317

∘

317, degrees.

Step-by-step explanation:

khan academy copy and paste

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