Question

Help Please Help...
Quadrilateral ABCD is inscribed in a circle. m A is 64°, m B is (6x + 4)°, and m C is (9x - 1)°. What is m D?
a.) 64°
b.) 82°
c.) 90°
d.) 98°
e.) 116°

Answer 1

@efco_chen

Answer 2

For a quadrilateral inscribed in a circle, opposite angles are supplementary, so
\[\angle A+\angle C=180^\circ\\
\angle B+\angle D=180^\circ\]
The first equation gives
\[64+9x-1=180~~\Rightarrow~~x=117\]
Plug this into the second equation:
\[6x+4+\angle D=180\]
and solve for \(\angle D\).

Answer 3

I don't understand

Answer 4

Sorry, that should be \(9x=117\), which gives \(x=13\).

If \(x=13\), what is \(6x\)?

Answer 5

that problem is that I don't know how to solve it :(

Answer 6

What's 6 times 13?

Answer 7

78

Answer 8

Right, so if \(x=13\), then \(6x=6\times13=78\).

Now, plug this into the equation containing \(\angle D\):
\[78+4+\angle D=180\]
Can you solve for \(\angle D\)?

Answer 9

soooo... 82? im sorry I just .. :( I suck at math :( im more of an artist :(

Answer 10

\[\begin{align*}78+4+\angle D&=180\\
82+\angle D&=180\\
82-82+\angle D&=180-82\\
0+\angle D&=180-82\\
\angle D&=?
\end{align*}\]

Answer 11

So is it 98?