Question

pls solve the question given in problem​

Answer 1

$\large\sf\underline{Given}$

ABCD is a cyclic quadrilateral in which :

• ∠A = ( 2x + 4 )°

• ∠B = ( x + 10 )°

• ∠C = ( 4y - 4 )°

• ∠D = ( 5y + 5 )°

$\large\sf\underline{To\:find}$

• Values of x and y .

• The measure of each angles .

$\large\sf\underline{Solution}$

We know ,

The opposite angles in a cyclic quadrilateral = $\small{\underline{\boxed{\mathrm\pink{180°}}}}$

So ,

$\sf\:∠A + ∠C = 180°$

$\sf➜\:(2x+4)° + (4y-4)° = 180°$

$\sf➜\:2x+4 +4y-4= 180$

$\sf➜\:2x+ 4y= 180$

$\sf➜\:2(x+ 2y)= 180$

$\sf➜\:x+ 2y= \frac{180}{2}$

$\sf➜\:x+2y=90\:----(I)$

And ,

$\sf\:∠B+ ∠D = 180°$

$\sf➜\:(x+10)° + (5y+5)° = 180°$

$\sf➜\:x+10 +5y+5= 180$

$\sf➜\:x+ 5y+15= 180$

$\sf➜\:x+ 5y= 180-15$

$\sf➜\:x+5y=165\:----(II)$

Now subtracting (I) from (II) we get :

$\sf\:(II)-(I)$

$\sf➙\:(x+5y)-(x+2y)=165-90$

$\sf➙\:x+5y-x-2y=165-90$

$\sf➙\:x-x+5y-2y=165-90$

$\sf➙\:3y=75$

$\sf➙\:y=\frac{75}{3}$

$\small{\underline{\boxed{\mathrm\purple{➤\:y=25°}}}}$

Substituting the value of y in equation (I) :

$\sf➙\:x+2y=90$

$\sf➙\:x+2 \times 25=90$

$\sf➙\:x+ 50=90$

$\sf➙\:x=90-50$

$\small{\underline{\boxed{\mathrm\purple{➤\:x=40°}}}}$

Now substituting the value of x and y in ∠A , ∠B , ∠C and ∠D we get :

• ∠A = ( 2x + 4 )°

$\sf⤐\:∠A =[(2 \times 40 )+ 4]°$

$\sf⤐\:∠A =[80 + 4]°$

$\small{\underline{\boxed{\mathrm\blue{➺\:∠A =\:84°}}}}$

• ∠B = ( x + 10 )°

$\sf⤐\:∠B =(40 + 10)°$

$\small{\underline{\boxed{\mathrm\blue{➺\:∠B =\:50°}}}}$

• ∠C = ( 4y - 4 )°

$\sf⤐\:∠C =[(4 \times 25)- 4]°$

$\sf⤐\:∠C =[100 - 4]°$

$\small{\underline{\boxed{\mathrm\blue{➺\:∠C =\:96°}}}}$

• ∠D = ( 5y + 5 )°

$\sf⤐\:∠D =[(5 \times 25)+5]°$

$\sf⤐\:∠D =[125 +5]°$

$\small{\underline{\boxed{\mathrm\blue{➺\:∠D =\:130°}}}}$

$\large\sf\underline{Verification\:of\:my\:answer}$

As opposite angles of a cyclic quadrilateral are supplementary , let's see if :

$\sf\:∠A + ∠C = 180°$

$\sf➤\:84° + 96° = 180°$

$\sf➤\:180° = 180°$

So LHS = RHS

Again ,

$\sf\:∠B + ∠D = 180°$

$\sf➤\:50° + 130° = 180°$

$\sf➤\:180° = 180°$

Hence LHS = RHS

Therefore the measure of all the angles are correct .

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