SOLUTION: Could you please help me with this question about restriction on x? Would greatly appreciate if you could please share the steps to solve it. <a rel=nofollow HREF="https://ibb.

Algebra ->  Triangles -> SOLUTION: Could you please help me with this question about restriction on x? Would greatly appreciate if you could please share the steps to solve it. <a rel=nofollow HREF="https://ibb.      Log On


   



Question 1186436: Could you please help me with this question about restriction on x? Would greatly appreciate if you could please share the steps to solve it.
IMG-4972-xrestrictions

Found 2 solutions by greenestamps, MathLover1:
Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


In the figure, let the triangle be ABC, with angle A 70 degrees and angle B 55 degrees; let CD be the segment from vertex C to base AB.

Imagine point D moving along base AB.

The farthest to the right point D can be is just to the left of vertex B; that means angle x has to be greater than 55 degrees.

The farthest to the left point D can be is just to the right of vertex A. Since the angle sum in triangle ACD is 180 degrees, x has to be less than 110 degrees. (70 plus x has to be less than 180 degrees.)

ANSWER: x is greater than 55 degrees and less than 110 degrees


Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
given:
first angle 70
second 55
then third angle is180-%2870%2B55%29=180-125=55
so, triangle is isosceles+because two of the legs are the same, and it is divided in two triangles (one with angles x and 70, the other with 55 and potion of other 55 angle)
then, angle x is an exterior angle for smaller triangle (with+55 and portion of other+55+angle)

by definition of exterior angle, x=55+portion of other+55+angle
so, conclusion is: x%3E55