Question

ABCD is a cyclic quadrilateral. If <ACD = 40°, <ADC = 80°. Find <CBD



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Answer 1

Step-by-step explanation:

Given that,

Angle dbc = 80

Angle bac = 40

We know that,

Angle in the same segment is equal.

So, \angle dac=\angle dbc∠dac=∠dbc

We need to calculate the angle bad

Using given data

\angle bad=\angle bac+\angle dac∠bad=∠bac+∠dac

Put the value into the formula

\angle bad=40+80∠bad=40+80

\angle bad=120∠bad=120

We need to calculate the value of angle bcd

According to figure,

\angle bcd+\angle bad=180∠bcd+∠bad=180

Put the value into the formula

\angle bcd=180-120∠bcd=180−120

\angle bcd=60∠bcd=60

I HOPE IT HELPS YOU

Hence, The value of angle bcd is 60°.

Answer 2

♣GIVEN♣

• $\begin{gathered}\sf \angle ACD = 40^{\circ}\\\end{gathered}$

• $\begin{gathered}\sf \angle ADC = 80^{\circ}\\\end{gathered}$

♣TO FIND♣

• $\begin{gathered}\sf Value\:of\:\angle CBD\\\end{gathered}$

♣DIAGRAM♣

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3) \qbezier(1.7,1.6)(1.7,2)(1.7,-1.6)\qbezier(-1.5,1.7)(-1.5,2)( -1.5,-1.7)\qbezier( - 1.5,1.7)(1.7,1.6)(1.7,1.6) \qbezier(-1.6, - 1.7)(1.7,-1.6)(1.7,-1.6)\qbezier( - 1.5,1.7)(1.7,-1.6)(1.7,-1.6)\qbezier(-1.6, -1.7)(1.7,1.6)(1.7,1.6)\qbezier(1.1, -1)(1,-1.5)(1,-1.6)\qbezier(-1,-1.7)( - 0.8,-0.7)(-1.5,-1)\put(-2,1.8){\bf A}\put(1.7,1.7){\bf B} \put(1.7, -1.9){\bf C} \put( - 2, - 1.9){\bf D}\put(0.43,- 1.4){ \bf {40}^{\circ} }\put( -1.2,-0.7){ \bf {80}^{\circ} } \end{picture}$

♣SOLUTION♣

• Let us f1st recall the properties of cyclic quadrilateral. In a cyclic quadrilateral sum of opp. angles is 180° and in a Circle angles made from the same segments are equal.

Here,

$: \sf \implies \: \: \: \: \: \: \angle ACD = \angle ABD = 40^{\circ} \qquad \bigg \lgroup Angles\:in\:same\: segment \bigg \rgroup$

$:\sf \implies \: \: \: \: \: \: \angle ABC + \angle ADC = 180^{\circ} \qquad \bigg \lgroup Opp. \:angles\:of\: cyclic\:quadr. \bigg \rgroup$

$\begin{gathered} \sf where {\begin{cases} \\ \sf\angle ADC = 80^{\circ} \\ \\ \end{cases}}\end{gathered}$

Substitute this value:-

$: \sf \implies \: \: \: \: \: \: \angle ABC + \angle ADC = 180^{\circ}$

$: \sf \implies \: \: \: \: \: \: \angle ABC + 80^{\circ} = 180^{\circ}$

$: \sf \implies\: \: \: \: \: \: \angle ABC= 180^{\circ} - 80^{\circ}$

$: \implies \: \: \: \: \: \: \underline{ \boxed{\sf \angle ABC=100^{\circ}} }$

Now,

$\sf \longrightarrow \: \: \: \: \: \angle ABC = \angle ABD + \angle CBD$

$\sf \longrightarrow \: \: \: \: \: \angle ABC - \angle ABD = \angle CBD$

$\sf where {\begin{cases}\sf\angle ABC = 100^{\circ}\\ \sf\angle ABD = 40^{\circ} \end{cases}}$

Substituting these values:-

$\sf \longrightarrow \: \: \: \: \: \angle CBD = \angle ABC - \angle ABD$

$\sf \longrightarrow \: \: \: \: \: \angle CBD = 100^{\circ} - 40^{\circ}$

$\sf \longrightarrow \: \: \: \: \: \underline{\boxed{ \sf{\angle CBD = 60^{\circ}} }}$

$\huge{ \boxed{\boxed{\bf \therefore{Value \:of\:\angle CBD=60^{\circ}}}}}$

♣ANSWER♣

• $\sf Value\:of\:\angle CBD =60^ { \circ}$