# Question #a9318

Jun 29, 2016

a) The sum of the angles of a triangle is ${180}^{\setminus} \circ$, then

$x + 3 x + 76 = 180$
$4 x = 180 - 76$
$x = \frac{104}{4} = 26$.

b) This is an isosceles triangle then the missing angle is $x$ because correspond to the same side. I apply again the rule of the angles writing

$x + 4 x + x = 180$
$6 x = 160$
$x = \frac{160}{6} = 26. \setminus \overline{6}$.

c) The sum of the angles of a quadrilateral is ${360}^{\setminus} \circ$, then

$x + 2 x + 4 x + 66 = 360$
$7 x = 360 - 66$
$x = \frac{294}{7} = 42$.

d) We solve as the problem a).

$x + 2 x + 57 = 180$
$3 x = 180 - 57$
$x = \frac{123}{3} = 41$

e) This is a parallelogram so the opposite angles are equal and the sum, as in the problem c) is ${360}^{\setminus} \circ$, then

$9 x + 6 x + 9 x + 6 x = 360$
$30 x = 360$
$x = \frac{360}{30} = 12$.

f) The problem is the same as problem a) but we have one external angle instead of the internal. We know that the sum between internal and external angle is ${180}^{\setminus} \circ$, then the internal angle is $180 - 108 = 72$. We can now proceed as in the exercise a)

$72 + 3 x + x - 4 = 180$
$4 x + 68 = 180$
$4 x = 180 - 68$
$x = \frac{112}{4} = 28$.

g) We solve as the problem c)

$2 x + x - 30 + x - 8 + 66 = 360$
$4 x + 28 = 360$
$4 x = 360 - 28$
$x = \frac{332}{4} = 83$.

h) This is a parallelogram so the opposite angles are has to be equal.

$2 x + 40 = 5 x + 25$
$3 x = 15$
$x = 3$.

i) We solve as the problem a)

$3 x + 30 + x + 60 + 6 x - 10 = 180$
$10 x + 80 = 180$
$10 x = 180 - 80$
$x = \frac{100}{10} = 10$.

j) This is an isosceles triangle, then the two angles has to be equal.

$2 x + 16 = 3 x + 9$
$x = 16 - 9 = 7$.