Question

If mBC = (9x-53) and mCD = (2x + 45) find mBAD

Answer 1

Answer:

$m\angle BAD=73^o$

Step-by-step explanation:

The picture of the question in the attached figure

step 1

Find the value of x

Let

O ----> the center of the circle

we know that

Triangle BOC≅Triangle COD

$m\angle BOC=arc\ BC$ ----> by central angle

$m\angle COD=arc\ CD$ ----> by central angle

$m\angle BOC=m\angle COD$

therefore

$arc\ BC=arc\ CD$

substitute the given values

$(9x-53)^o=(2x+45)^o$

solve for x

$9x-2x=45+53\\7x=98\\x=14$

step 2

Find the measure of angle BAD

we know that

The inscribed angle is half that of the arc it comprises.

so

$m\angle BAD=\frac{1}{2} [arc\ BC+arc\ CD]$

$arc\ BC=9(14)-53=73^o$

$arc\ CD=2(14)+45=73^o$

substitute

$m\angle BAD=\frac{1}{2} [73^o+73^o]=73^o$