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3 months ago

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

Mathematics

1 answer:

babunello [11.8K]3 months ago

7 0

Answer:

$m\angle BAD=73^o$

Step-by-step explanation:

The diagram related to the question is in the attached figure.

Step 1

Determine the value of x

Let:

O ----> denote the center of the circle

It is known that:

Triangle BOC is congruent to Triangle COD

$m\angle BOC=arc\ BC$ ----> due to the central angle

$m\angle COD=arc\ CD$ ----> according to the central angle

$m\angle BOC=m\angle COD$

Thus:

$arc\ BC=arc\ CD$

Insert the provided values

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

Solve for x

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

Step 2

Calculate the measure of angle BAD

It is known that:

The inscribed angle measures half that of the arc it subtends.

Thus:

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

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

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

Insert values

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

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