Question

find find the four angle of cyclic quadrilateral ABCD in which angle A equal to 2 x - 1 and angle b y + 5 and angle c 2 y + 15 and angle d 4 x minus 7.​

Answer 1

Answer:

⇒∠A=(2x−1)=2×33−1=65°

⇒∠B=(y+5)=50+5=55°

⇒∠C=(2y+15)=2×50+15=115°

⇒∠D=(4x−7)=4×33−7=125°

Step-by-step explanation:

⇒∠A=(2x−1)°

⇒∠B=(y+5)°

⇒∠C=(2y+15)°

⇒∠D=(4x−7)°

The sum of the opposite angle of a cyclic quadrilateral is 180°

∴∠A+∠C=180°

∠C+∠D=180°

⇒(2x−1)+(2y+15)=180°

⇒2x+2y=166°

⇒x+y=83 .....(1)

⇒(y+5)+(4x−7)=180°

→4x+y=182

....(2)

subtract eq1 and eq2

⇒−3x=−99⇒x=33

put x=33 in eq1

⇒33+y=83⇒y=50

Hence,

⇒∠A=(2x−1)=2×33−1=65°

⇒∠B=(y+5)=50+5=55°

⇒∠C=(2y+15)=2×50+15=115°

⇒∠D=(4x−7)=4×33−7=125°

hope it helps.

please mark me BRAINIEST (: