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The smallest angle of a triangle is equal to two thirds of the smallest angle of a quadrilateral. The ratio between the angles of the quadrilateral is $3:4:5:6.$ The largest angle of the triangle is twice its smallest angle. What is the sum, in degrees, of the second largest angle of the triangle and the largest angle of the quadrilateral?
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Sum of the angles of a quadrilateral $= 360$

Therefore, $3x + 4x + 5x + 6x = 360$
$\implies 18x = 360$
$\implies x = 360/18= 20$

Therefore, smallest angle of quadrilateral $= 3×20 = 60$

Largest angle of quadrilateral $= 6×20 = 120$

Therefore, smallest angle of triangle $= 60×2/3=40$

Largest angle of triangle $= 2×40 = 80$

Therefore, third angle of triangle $= 180−40−80=60$

Required sum $= 60 +120 = 180$
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Clearly, the angles of quadrilateral are 60, 80, 100, 120.

That makes smallest angle of triangle 60*(2/3) = 40.

That makes largest angle of triangle 40*(2) = 80.

so third angle of triangle = 60.

The required value is = 120 + 60 =180

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The ratio between the angles of the quadrilateral is 3:4:5:6

that means smallest :3x , largest:6x.

consider a triangle with angle a , b ,c 

a= smallest , b= largest, c= second largest

as per given info:

a= 2/3*(3x) = 2x

b= 2*a = 2*2x =4x

i.e c = 180-(a+b) = 180- 6x

we have to find second largest angle of the triangle and the largest angle of the quadrilateral:

== c (second largest angle of the triangle) + 6x( largest angle of the quadrilateral )

= (180 – 6x) + 6x= 180

Answer:

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