Question

in the given figure , AB CD . if angle BAC = (3x + 15)° and angle ACD = (2x + 45)° , find the value of x , also , find the measures of angle BAC and angle ACD​

Answer 1

Answer:

$\red{ \boxed{ \sf\red{Given:-}}}$

• ACD = (2x + 45)°
• BAC = (3x + 15)°

$\boxed{ \red{ \sf \: To \: Find : - }}$

• measures of angle BAC and angle ACD

$\pink{ \sf \: Solution:-}$

$\sf \: 3x + 15 = 2x + 45$

$\sf \: 3x - 2x = 45 - 15$

$\pink{ \sf \frak x = 35 {}^{0} }$

$\sf \: BAC= 3x+15$

$\sf \: 3(30) + 15$

$\red{ \sf \: = 105 {}^{0} }$

$\sf \: ACD=2x+45$

$\sf \: 2(30) + 45$

$\red{ \sf \: \frak{105 {}^{0} }}$

BAC=105°

ACD= 105°

Hence, This is Answer

Answer 2

3x+15+2x+45=180

5x+60=180

5x=180-60

5x=120

x=120/5x=24

/_ ACD=(2X+45)

=2(24)+45

48+45

=93

/_ BAC= (3X+15)

=3(24)+15

=72+15

=87

X=24

/_ACD=93

/_BAC=87