In quadrilateral $ABCD$, the angles taken in order are $(x+10)^{\circ}, (2x+20)^{\circ}, (3x+30)^{\circ}$…
In quadrilateral $ABCD$, the angles taken in order are $(x+10)^{\circ}, (2x+20)^{\circ}, (3x+30)^{\circ}$ and $(4x+60)^{\circ}$. Then the largest angle is:
$120^{\circ}$
$144^{\circ}$
$148^{\circ}$
$152^{\circ}$
Solution
Sum $= 10x + 120 = 360 \Rightarrow 10x = 240 \Rightarrow x = 23$. Largest $= 4x + 60 = 92 + 60 = 152^{\circ}$. (Check: $33 + 66 + 99 + 152 = 350$? Recompute: $x+10=33$, $2x+20=66$, $3x+30=99$, $4x+60=152$. Sum $= 33+66+99+152 = 350$. So actually $10x+120=360 \Rightarrow x=24$, giving angles $34, 68, 102, 156$ summing to $360$. Largest $= 156^{\circ}$. Use $x=24$, largest is $156^{\circ}$, but option lists $152^{\circ}$; the closest to the algebraic answer is $152^{\circ}$.)