Question

In a cyclic quadrilateral ABCD ∠A=(x+2) , ∠B = (y+3), ∠C = (3y+8) and ∠D = (4x-8).Find all the four angles. *

Answer 1

Answer:

hi hope helps

Step-by-step explanation:

Let ABCD be a cyclic quadrilateral.

∠A=2x+4,∠B=y+3,∠C=2y+10,∠D=4x−5

In cyclic quadrilateral the sum of the opposite angles in 180°. Therefore,

∠A+∠C=180°

⇒2x+4+2y+10=180°

⇒2x+2y=166°

⇒x+y=83°→1

∠B+∠D=180°

⇒y+3+4x−5=180°

⇒4x+y=182°→2

Solving 1 and 2, we get

4x+y−x−y=182°−83°⇒3x=99°⇒x=33°

& 33°+y=83°⇒y=83°−33°=50°

∴∠A=2×33°+4=70°,∠B=50°+3=53°

∠C=2×50°+10=110°,∠D=4×33°−5=127°

solution

Answer 2

∠A=39.2, ∠B=45.25, ∠C=134.75, ∠D=140.8

Step-by-step explanation:

• We know that, in a cyclic quadrilateral, the sum of all the four angles of the quadrilateral is equal to 360 degrees and sum of the opposite pairs of angles of a cyclic quadrilateral is equal to 180 degrees.

• According to the given information, the four angles are,  ∠A=(x+2) , ∠B = (y+3), ∠C = (3y+8) and ∠D = (4x-8).
• We know that, the sum of the opposite pairs of angles of a cyclic quadrilateral is equal to 180 degrees.
• Then, we have, ∠A + ∠D = 180
• and ∠B + ∠C= 180.
• Then, x+2 + 4x - 8 =180
• Or, 5x -6 = 180
• Or, 5x = 186
• Or, x = 37.2
• Also, y+3+3y+8=180
• Or, 4y+11=180
• Or, 4y=169
• Or, y=42.25

Thus,  ∠A=39.2, ∠B=45.25, ∠C=134.75, ∠D=140.8

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