Question

In the following figure, find the value of x:​

Answer 1

Answer:

x = 31

Step-by-step explanation:

first Lets focus on ADC. we can observe that angle DAC+ angle ADC = 180 - 56(because sum of all angles =  180)

AC  and DC are equal so angle DAC and angle ADC are equal( because angles opposite to equal sides are equal)

lets assume the angles to be y

so y + y = 124

2y = 124

y= 62

Lets focus on triangle ABD

angle BDA = 180- 62  ( because it is straight line)

we can observe that BD and DA are equal so x and angle ABD are equal

so x+x+118 = 180

2x = 62

x = 31

Really hope this helped

Answer 2

$\huge\tt{Answer:-}$

$\bf{Given:-}$

• $\sf{AD = BD}$.
• $\sf {AC = DC}$.
• $\sf{\angle ACD = {56}^{\degree}}$

$\bf{To \ Find:-}$

Value of $\sf{ {x}^{\degree}}$ here.

__________...

Consider $\sf{\triangle ADC}$,

Here,

$\sf{\triangle ADC}$ is an Isosceles Triangle. [∵ 2 opposite sides are equal.]

Let one of the same angle be = $\sf{ {N}^{\degree} }$.

$\sf{ \therefore {N}^{\degree} + {N}^{\degree} + {56}^{\degree} = {180}^{\degree}}$

(Since sum of int. angles are = 180°.)

$\sf{ \implies 2{N}^{\degree} = {180}^{\degree} - {56}^{\degree} }$

$\sf{\implies {N}^{\degree} = \frac{{124}^{\degree} }{2} }$

$\sf{\implies {N}^{\degree} = {62}^{\degree}}$

____________...

Finding the value of angle BDA:-

$\sf{ \angle BDC = \angle BDA + \angle ADC}$

$\sf{\implies \angle BDA + {62}^{\degree} = {180}^{\degree}}$ (Linear Pair)

$\sf{\implies \angle BDA = {180}^{\degree} - {62}^{\degree}}$

$\sf{\implies \angle BDA = {118}^{\degree}}$

____________...

Finding the value of (x°) :-

$\sf{\triangle BDA = 2{x}^{\degree} + \angle BDA = {180}^{\degree}}$

$\sf{\implies 2{x}^{\degree} + {118}^{\degree} = {180}^{\degree}}$

$\sf{\implies 2{x}^{\degree} = {180}^{\degree} - {118}^{\degree}}$

$\sf{\implies 2{x}^{\degree} = {62}^{\degree}}$

$\sf {\implies {x}^{\degree} = \frac{{62}^{\degree}}{2}}$

$\boxed{\sf {\implies {x}^{\degree} = {31}^{\degree}}}$ $... (Ans.)$