Question

ABCD is a cyclic quadrilateral . If ∠A= 2x° , ∠B=6y+10° , ∠C= 2x+y° and ∠D= x + 10° find the smallest angle.
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Answer 1

☯DIAGRAM :

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1,1)(1,1)(6,1)\put(0.4,0.5){\bf D}\qbezier(1,1)(1,1)(1.6,4)\put(6.2,0.5){\bf C}\qbezier(1.6,4)(1.6,4)(6.6,4)\put(1,4){\bf A}\qbezier(6,1)(6,1)(6.6,4)\put(6.9,3.8){\bf B}\end{picture}$

☯GIVEN :

• Angles of the quadrilateral are (2x)°,(6y+10)°, (2x+y)°and (x+10)°.

☯TO FIND :

• The smallest angle.

☯SOLUTION :

As we know that,

• $\boxed{\bf {Sum\:of\:all\:angles=360^{\circ}}}$

Substituting the given values :

$:\longrightarrow \sf 2x+(6y+10)+(2x+y)+(x+10)=360$

$:\longrightarrow \sf 2x+6y+10+2x+y+x+10=360$

$:\longrightarrow \sf 2x+2x+x+6y+y+10+10=360$

$:\longrightarrow \sf 5x+7y+20=360$

$:\longrightarrow \sf 5x+7y=360-20$

$:\longrightarrow\tt 5x+7y=340\: ....(1)$

• Again ,

$:\longrightarrow \sf \angle A+ \angle B=180^ {\circ}$

$:\longrightarrow\sf 2x+6y+10=180$

$:\longrightarrow\sf 2x+6y=180-10$

$:\longrightarrow \tt 2x+6y=170\: .... (2)$

• Now use elimination method by multiplying 2 in eq (1) and multiplying 5 in eq (2) we get,

$\tt 10x+14y=680\: ....(3)$

$\tt 10x+30y=850\: ....(4)$

_____________________________

Subtract eq (3) from eq (4) :

$:\longrightarrow \tt 16y=170$

$:\longrightarrow \tt {y = \cancel{\dfrac{170}{16}} }$

$:\longrightarrow{\underline{\boxed{\bf{y=10.625}}}}$

Substitute the value of y in eq (3) :

$:\longrightarrow \sf 10x+14 \times10.625=680$

$:\longrightarrow \sf 10x+148.75=680$

$:\longrightarrow \sf 10x=680 - 148.75$

$:\longrightarrow \sf 10x=531.25$

$:\longrightarrow \sf x= \cancel{\dfrac{531.25}{10}}$

$: \longrightarrow{\underline{\boxed{\bf {x=53.125}}}}$

_______________________

Hence , the angles are :

• (2x)° = (2×53.125)°= 106.25°

• (6y+10)°=(6×10.625+10)°=73.75°

• (2x+y)°=(2×53.125+ 10.625)°=106.25+10.625=116.875°

• (x+10)°=(53.125+10)°= 63.125°

$\huge {\therefore }$ $\bf {\angle D}$ is the smallest angle.