Question

B, C and D are points on a circle, centre O.
AB and AD are tangents to the circle.
Angle BCD=x
Find an expression for the size of angle BAD in terms of x.
Give your answer in its simplest form.
Give reasons for each stage of your working.

Answer 1

Answer:

$(180 - 2x)\degree$

Step-by-step explanation:

In circle with center O, $m\angle BCD = x\degree ... (given)$

By inscribed angle theorem:

$m\angle BCD= \frac{1}{2} \times m\widehat {BD} \\

\therefore x\degree = \frac{1}{2} \times m\widehat {BD} \\

\therefore \widehat {BD} = 2x\degree \\

\because m\widehat {BCD}+m\widehat {BD}= 360\degree \\(By\: arc\: sum\: postulate\: of\: a \: circle) \\

\therefore m\widehat {BCD}+2x\degree = 360\degree \\

\therefore m\widehat {BCD} = 360\degree - 2x\degree \\\\

\because m\angle BAD = \frac{1}{2}(m\widehat {BCD} - m\widehat {BD}) \\\\

\therefore m\angle BAD = \frac{1}{2}(360\degree - 2x\degree - 2x\degree) \\\\

\therefore m\angle BAD = \frac{1}{2}(360\degree - 4x\degree ) \\\\

\therefore m\angle BAD = \frac{1}{2}\times 2(180\degree - 2x\degree \\\\

\red {\boxed {\bold {\therefore m\angle BAD =(180- 2x)\degree}}}$

Answer 2

Answer in its simplest form. = ∠BAD  = 180° - $\bold {2x}$

B, C and D are points on a circle, center O.

AB and AD are tangents to the circle

We have to find an expression for the size of ∠BAD in terms of x.

∠BCD = x ( Given )

Expression for the size of ∠BAD in terms of x can be derived from the following steps

Center angle is twice the angle inscribed at the arc of a circle .

∠BCD =(1/2) ∠ BOD

$x =$ (1/2) ∠ BOD

$\bold{2x}$ = ∠ BOD ........ (1)

So in quadrilateral BODA

∠BOD + ∠ BAD  = 180° ........(2)  ( By two tangent theorem)

Since ∠ ODA = 90 °

         ∠ OBD = 90 °

From Equation (1) and (2)

∠BAD  = 180° - $\bold {2x}$

Answer in its simplest form is  given by the expression given as follows ∠BAD  = 180° - $\bold {2x}$

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