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I usually don't do this but I'm sick and missed class. Any hepl would be appreciate for this problem!

 

A circle is circumscribed around $ABCD$ as follows:

[asy]
pair pA, pB, pC, pD, pO;
pO = (0, 0);
pA = pO + dir(-40);
pB = pO + dir(40);
pC = pO + dir(130);
pD = pO + dir(190);
draw(pA--pB--pC--pA);
draw(pA--pD--pC--pA);
label("$A$", pA, SE);
label("$B$", pB, NE);
label("$C$", pC, NW);
label("$D$", pD, SW);
draw(circle(pO, 1));
label("$30^\circ$", pA + dir(150) * .45);
label("$40^\circ$", pC + dir(-20) * .35);
[/asy]

 

How many degrees are in $\angle CAB + \angle ACD$?

 Feb 9, 2020
 #1
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Angle CAB + angle ACD = 30 + 40 + 90 = 160 degrees.

 Feb 9, 2020
 #3
avatar+2850 
0

where did you get the 90?

CalculatorUser  Feb 9, 2020
 #2
avatar+2850 
+1

You are lucky I use AoPS and was able to render your image :/

Angle DCB + Angle DAB = 180

Lets call ACD \(x\)

and lets call CAB \(y\)

 

\(x+y+70=180\)

 

Now.

 

Can you figure out the answer? (simple algebra!)

 Feb 9, 2020

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