The two adjacent sides of a cyclic quadrilateral is 2 and 5 and the angle between them is 60∘. If the area of the quadrilateral is 43 , find the remaining two sides.

Let the adjacent sides of the cyclic quadrilateral be AB and BC respectively.Therefore, it is given that,⇒AB=2⇒BC=5⇒Angle between AB and BC ∠ABC=60∘Now, we know, opposite angles of a quadrilateral =180∘.Therefore, according to the diagram,⇒∠ABC+∠ADC=180∘⇒60∘+∠ADC=180∘⇒∠ADC=180∘-60∘⇒∠ADC=120∘ Let the other two adjacent sides be CD=a and AD=b.Now,We know, that according to the sine rule, the area of the triangle is equal to,⇒Δ=12×product of the lengths of two sides×sin (angle between the sides) [equation I]Therefore,In Δ ABC,⇒Δ1=12×AB×BC×sin∠ABC [from equation I]Putting the values,⇒Δ1=12×2×5×sin60∘⇒Δ1=5×32⇒Δ1=532 [equation II] Similarly,In Δ ADC,⇒Δ2=12×CD×AD×sin∠ADC [from equation I]Putting the values,⇒Δ2=12×a×b×sin120∘⇒Δ2=ab2×32⇒Δ2=ab34 [equation III] ⇒Area of the quadrilateral =Δ1+Δ2⇒Aq=532+ab34 [From equation II and III]⇒43-532=ab34 [Area of the quadrilateral is given]⇒83-532=ab34⇒332=ab34⇒ab=33×423⇒ab=6 [equation IV] Now, we also know the cosine rule, that the relation of the Δ ABC with given three sides AB, BC and AC and the angle between AB and BC=∠ABC is given by,⇒AC2=AB2+BC2-2×AB×BC×cos∠ABC [equation V]Similarly, in Δ ADC,⇒AC2=CD2+AD2-2×CD×AD×cos∠ADC [equation VI] Therefore, from equation V and VI, we can write,⇒AB2+BC2-2×AB×BC×cos∠ABC=CD2+AD2-2×CD×AD×cos∠ADC⇒22+52-2×2×5×cos60∘=a2+b2-2a×b×cos120∘⇒29-20×12=a2+b2-2a×b×-12⇒29-10=a2+b2+ab⇒19=a2+b2+6 [From equation IV]⇒19-6=a2+b2⇒a2+b2=13 [equation VII] Now we will be forming a quadratic equation using a2+b2=13 and a2b2=36, where, a2, b2 are the roots of the equation forming with x.⇒x2-a2+b2x+a2b2=0⇒x2-13x+36=0⇒x2-(9+4)x+36=0⇒x2-9x-4x+36=0⇒x(x-9)-4(x-9)=0⇒(x-9)(x-4)=0Either,⇒x-9=0⇒x=9⇒a2=9⇒a=3Or,⇒x-4=0⇒x=4⇒a2=4⇒a=2 Similarly,⇒b=3, 2 Therefore, the length of the other two sides are:⇒CD=a=3, 2⇒AD=3, 2

The two adjacent sides of a cyclic quadrilateral is 2 and 5 and the angle between them is 60∘. If the area of the quadrilateral is 43 , find the remaining two sides.

Let the adjacent sides of the cyclic quadrilateral be AB and BC respectively.Therefore, it is given that,⇒AB=2⇒BC=5⇒Angle between AB and BC ∠ABC=60∘Now, we know, opposite angles of a quadrilateral =180∘.Therefore, according to the diagram,⇒∠ABC+∠ADC=180∘⇒60∘+∠ADC=180∘⇒∠ADC=180∘-60∘⇒∠ADC=120∘ Let the other two adjacent sides be CD=a and AD=b.Now,We know, that according to the sine rule, the area of the triangle is equal to,⇒Δ=12×product of the lengths of two sides×sin (angle between the sides) [equation I]Therefore,In Δ ABC,⇒Δ1=12×AB×BC×sin∠ABC [from equation I]Putting the values,⇒Δ1=12×2×5×sin60∘⇒Δ1=5×32⇒Δ1=532 [equation II] Similarly,In Δ ADC,⇒Δ2=12×CD×AD×sin∠ADC [from equation I]Putting the values,⇒Δ2=12×a×b×sin120∘⇒Δ2=ab2×32⇒Δ2=ab34 [equation III] ⇒Area of the quadrilateral =Δ1+Δ2⇒Aq=532+ab34 [From equation II and III]⇒43-532=ab34 [Area of the quadrilateral is given]⇒83-532=ab34⇒332=ab34⇒ab=33×423⇒ab=6 [equation IV] Now, we also know the cosine rule, that the relation of the Δ ABC with given three sides AB, BC and AC and the angle between AB and BC=∠ABC is given by,⇒AC2=AB2+BC2-2×AB×BC×cos∠ABC [equation V]Similarly, in Δ ADC,⇒AC2=CD2+AD2-2×CD×AD×cos∠ADC [equation VI] Therefore, from equation V and VI, we can write,⇒AB2+BC2-2×AB×BC×cos∠ABC=CD2+AD2-2×CD×AD×cos∠ADC⇒22+52-2×2×5×cos60∘=a2+b2-2a×b×cos120∘⇒29-20×12=a2+b2-2a×b×-12⇒29-10=a2+b2+ab⇒19=a2+b2+6 [From equation IV]⇒19-6=a2+b2⇒a2+b2=13 [equation VII] Now we will be forming a quadratic equation using a2+b2=13 and a2b2=36, where, a2, b2 are the roots of the equation forming with x.⇒x2-a2+b2x+a2b2=0⇒x2-13x+36=0⇒x2-(9+4)x+36=0⇒x2-9x-4x+36=0⇒x(x-9)-4(x-9)=0⇒(x-9)(x-4)=0Either,⇒x-9=0⇒x=9⇒a2=9⇒a=3Or,⇒x-4=0⇒x=4⇒a2=4⇒a=2 Similarly,⇒b=3, 2 Therefore, the length of the other two sides are:⇒CD=a=3, 2⇒AD=3, 2

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The two adjacent sides of a cyclic quadrilateral is $2$ and $5$ and the angle between them is $60^{\circ}$ . If the area of the quadrilateral is $4 \sqrt{3}$ , find the remaining two sides.

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Detailed Solution

Let the adjacent sides of the cyclic quadrilateral be

AB

and

BC

respectively.
Therefore, it is given that,

\Rightarrow AB = 2

\Rightarrow BC = 5

\Rightarrow

Angle between

AB

and

BC (∠ ABC) = 60^{\circ}

Now, we know, opposite angles of a quadrilateral

= 180^{\circ}

.
Therefore, according to the diagram,

\Rightarrow ∠ ABC + ∠ ADC = 180^{\circ}

\Rightarrow 60^{\circ} + ∠ ADC = 180^{\circ}

\Rightarrow ∠ ADC = 180^{\circ} - 60^{\circ}

\Rightarrow ∠ ADC = 120^{\circ}

Let the other two adjacent sides be

CD = a

and

AD = b

.
Now,
We know, that according to the sine rule, the area of the triangle is equal to,

\Rightarrow Δ = \frac{1}{2} \times product of the lengths of two sides \times \sin (angle between the sides)

[equation I]
Therefore,
In

Δ ABC

\Rightarrow Δ_{1} = \frac{1}{2} \times AB \times BC \times \sin (∠ ABC)

[from equation I]
Putting the values,

\Rightarrow Δ_{1} = \frac{1}{2} \times 2 \times 5 \times \sin (60^{\circ})

\Rightarrow Δ_{1} = 5 \times \frac{\sqrt{3}}{2}

\Rightarrow Δ_{1} = \frac{5 \sqrt{3}}{2}

[equation II]
Similarly,
In

Δ ADC

{\Rightarrow Δ}_{2} = \frac{1}{2} \times CD \times AD \times \sin (∠ ADC)

[from equation I]
Putting the values,

\Rightarrow Δ_{2} = \frac{1}{2} \times a \times b \times \sin (120^{\circ})

\Rightarrow Δ_{2} = \frac{ab}{2} \times \frac{\sqrt{3}}{2}

\Rightarrow Δ_{2} = \frac{ab \sqrt{3}}{4}

[equation III]

\Rightarrow

Area of the quadrilateral

= Δ_{1} + Δ_{2}

\Rightarrow A_{q} = \frac{5 \sqrt{3}}{2} + \frac{ab \sqrt{3}}{4}

[From equation II and III]

\Rightarrow 4 \sqrt{3} - \frac{5 \sqrt{3}}{2} = \frac{ab \sqrt{3}}{4}

[Area of the quadrilateral is given]

\Rightarrow \frac{8 \sqrt{3} - 5 \sqrt{3}}{2} = \frac{ab \sqrt{3}}{4}

\Rightarrow \frac{3 \sqrt{3}}{2} = \frac{ab \sqrt{3}}{4}

\Rightarrow ab = \frac{3 \sqrt{3} \times 4}{2 \sqrt{3}}

\Rightarrow ab = 6

[equation IV]
Now, we also know the cosine rule, that the relation of the

Δ ABC

with given three sides

AB, BC and AC

and the angle between

AB

and

BC = ∠ ABC

is given by,

\Rightarrow {AC}^{2} = {AB}^{2} + {BC}^{2} - 2 \times AB \times BC \times \cos (∠ ABC)

[equation V]
Similarly, in

Δ ADC

\Rightarrow {AC}^{2} = {CD}^{2} + {AD}^{2} - 2 \times CD \times AD \times \cos (∠ ADC)

[equation VI]
Therefore, from equation V and VI, we can write,

\Rightarrow {AB}^{2} + {BC}^{2} - 2 \times AB \times BC \times \cos (∠ ABC) = {CD}^{2} + {AD}^{2} - 2 \times CD \times AD \times \cos (∠ ADC)

\Rightarrow 2^{2} + 5^{2} - 2 \times 2 \times 5 \times \cos (60^{\circ}) = a^{2} + b^{2} - 2 a \times b \times \cos (120^{\circ})

\Rightarrow 29 - 20 \times \frac{1}{2} = a^{2} + b^{2} - 2 a \times b \times (- \frac{1}{2})

\Rightarrow 29 - 10 = a^{2} + b^{2} + ab

\Rightarrow 19 = a^{2} + b^{2} + 6

[From equation IV]

\Rightarrow 19 - 6 = a^{2} + b^{2}

\Rightarrow a^{2} + b^{2} = 13

[equation VII]
Now we will be forming a quadratic equation using

a^{2} + b^{2} = 13

and

a^{2} b^{2} = 36

, where,

a^{2}, b^{2}

are the roots of the equation forming with

x

\Rightarrow x^{2} - (a^{2} + b^{2}) x + a^{2} b^{2} = 0

\Rightarrow x^{2} - 13 x + 36 = 0

\Rightarrow x^{2} - (9 + 4) x + 36 = 0

\Rightarrow x^{2} - 9 x - 4 x + 36 = 0

\Rightarrow x (x - 9) - 4 (x - 9) = 0

\Rightarrow (x - 9) (x - 4) = 0

Either,

\Rightarrow x - 9 = 0

\Rightarrow x = 9

\Rightarrow a^{2} = 9

\Rightarrow a = 3

Or,

\Rightarrow x - 4 = 0

\Rightarrow x = 4

\Rightarrow a^{2} = 4

\Rightarrow a = 2

Similarly,

\Rightarrow b = 3, 2

Therefore, the length of the other two sides are:

\Rightarrow CD = a = 3, 2

\Rightarrow AD = 3, 2

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The two adjacent sides of a cyclic quadrilateral is 2 and 5 and the angle between them is 60∘. If the area of the quadrilateral is 43 , find the remaining two sides.

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The two adjacent sides of a cyclic quadrilateral is $2$ and $5$ and the angle between them is $60^{\circ}$ . If the area of the quadrilateral is $4 \sqrt{3}$ , find the remaining two sides.