# how do you solve this triangle equation [GEOMETRY]?

i don't really know that its called but its somewhat in the fields of triangle. how can i solve this? i was absent for a couple of months due to surgery & recovery, so i didn't quite catch up. and unfortunately, geometry is the only subject i always fail on. please teach me how to solve this:

---> http://pichut.eu/x/ssas.jpg <--- sample equations

Relevance

for question 14:

You have a triangle whose base rests on a line, which is parallel to another line that is intersecting the other vertex of the triangle. You are given two angles of the triangle. (<3 and <2). You want to figure out <1,<4 and <5.

<3=55 deg.

<2=74 deg.

<4 is the other angle of the triangle. Since the sum of all the angles on a triangle add up to 180 deg. 180-55-74 should give us <4. <4=51

<2 and <1 are supplementary angles(a line is 180), so they have to add up to 180 deg. 180-74=106 deg. <1=106 deg.

<5 =<4 due to laws regarding a line intersecting two parallel lines. <5=51

20) The triangle ABC is compose of 3 triangles: ABD,ADE,AEC.

<1=2x

<3=x

Find the measure of angle B in terms of x?

First find angle 5 which is the supplementary angle of <1. 2x+<5=180; <5=180-2x

Since you have 2 angles of triangle ABD. You can figure out the third angle.

180-2x+x+<ABD=180

-x+<ABD=0

<ABD=x

<B is complementary to <ABD, so <B+x=360,

<B=360-x

Source(s): texbook