Question

Question #a9318

Answer 1

a) The sum of the angles of a triangle is `180^\circ`, then

`x+3x+76=180`
`4x=180-76`
`x=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+4x+x=180`
`6x=160`
`x=160/6=26.\bar{6}`.

c) The sum of the angles of a quadrilateral is `360^\circ`, then

`x+2x+4x+66=360`
`7x=360-66`
`x=294/7=42`.

d) We solve as the problem a).

`x+2x+57=180`
`3x=180-57`
`x=123/3=41`

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

`9x+6x+9x+6x=360`
`30x=360`
`x=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^\circ`, then the internal angle is `180-108=72`. We can now proceed as in the exercise a)

`72+3x+x-4=180`
`4x+68=180`
`4x=180-68`
`x=112/4=28`.

g) We solve as the problem c)

`2x+x-30+x-8+66=360`
`4x+28=360`
`4x=360-28`
`x=332/4=83`.

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

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

i) We solve as the problem a)

`3x+30+x+60+6x-10=180`
`10x+80=180`
`10x=180-80`
`x=100/10=10`.

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

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