Question

in a cyclic quadrilateral PQRS angle P=(2x+4)° , angle Q=(y+3)°, angle R=(2y+10)° and angle S=(4x-5)°,then find smallest angle
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Answer 1

As PQRS is cyclic quadrilateral

p + r = 180

q + s = 180

Now

P + R = 180

2x+4 + 2y + 10 = 180

2(x+y) = 166

x + y = 83.......(1)

Now

Q + S = 180

y + 3 + 4x-5 = 180

4x + y = 182.....(2)

Subtracting (1) from (2)

4x+y-(x+y) = 182 - 83

3x = 99

x = 33

y = 83 - x

= 83 - 33 = 50

Now

P = 2(33)+4 = 70

Q = 50 + 3 = 53

R = 2(50)+10

= 110

S = 4(33)-5 = 127

Hence Q is the smallest angle

Hope you gotta

Sahil Thakur

Answer 2

Answer:

∠P=70

∠Q=53

∠R=110

∠S=127

Step-by-step explanation:

In a cyclic quadrilateral sum of opposite angles is 180

∴ ∠R+∠P=180

   ∠Q+∠S=180

SINCE ∠R+∠P=180,

2X+4+2Y+10 = 180

2(X+Y) = 166

X+Y = 83------------- 1

similarly

∠Q+∠S=180

SO,

Y+2+4X-5=180

4X+Y=182------------2

SUBTRACT 1 FROM 2

WE GET  X= 33

SUBSTITUTE X=33 IN 2

33+Y=83

∴Y= 50

NOW SUBSTITUTE THE VALUES

∠P= 2×33+4 = 70

∠Q=50+3=53

∠R=2×50+10= 100 +10=110

∠S= 4×33-5 =  127

HOPE IT WILL HELP YOU

THANKYOU