Question

С
6. From the following figure find,
(i) X
(ii) angle ABC
(iii) angle ACD
​

Answer 1

Answer:-

Tip:-

Sum of interior angles of a $△$ is equal to $\sf{180°}$.

_________...

In $\sf{△ ABC}$,

$\sf{\angle BAC + \angle ABC + \angle ACB = {180}^{\degree}}$

$\sf{\implies x + 4x + 3x = 180°}$

$\boxed{\sf{\implies x = 22.5°}}$ (Answer 1)

So,

$\boxed{\sf{2. \angle ABC = 4x = 4(22.5°) = 90°}}$ (Answer 2)

In $\sf{△ ADC }$,

$\sf{\angle ADC + \angle CAD + \angle ACD = 180°}$

$\sf{\implies 4x + 48° + \angle ACD = 180°}$

By putting the value of 'x' and ending it,

$\boxed{\sf{\therefore \angle ACD = 42°}}$ (Answer 3)

Answer 2

Answer:

Q6.)=As per our question,

(i)=Value of x is 22.5°

(ii)=Value of angle(ABC) is 90°

(iii)=Value of angle(ACD) is 42°.

Step-by-step explanation:

Here, As per our given question,

ABCD is a quadrilateral in AC is its given diagonal,

(i)=So, In triangle ABC, we have,

=3x+4x+x=180° (By Angle-sum property of a triangle)

=8x=180°

=x=180°

=x=180°/8

=x=22.5°

So, The value of x=22.5°

(ii)=As we have value of x,

So,=angle(ABC)=4×22.5°

=90°

So, The value of angle(ABC)=90°.

(iii)=Now, In triangle ACD, we have,

=4x+48°+angle(ACD)=180° (By Angle-sum property of a triangle)

=4×22.5°+48°+angle(ACD)=180°

=90°+48°angle(ACD)=180°

=angle(ACD)=180°-138°

=angle(ACD)=42°

Hence, The value of angle(ACD)=42°.

Thank you.