Help with finding X in a triangle.

[Alpha6]

A little assistance please.

Since C is supplementary I tried: 5x- 2 + x = 180. Which left me with: 6x = 182 --> x = 30.3

When I plug in 30.3 for "x" it does NOT give me 180 degrees for the total triangle.

Help please.

 

[Alpha6]

No responses?

Never mind, I got it now.

 

[stapel]

No responses?

Never mind...

Sorry, but whoever told you that this was an on-demand service with paid staffers waiting to provide immediate replies (such as within the forty-five minute limit) was very much mistaken. This remains a free forum (not a chat-room or other "live" interface) where volunteers surf by to provide assistance as they're able, just like you saw in the "Read Before Posting" thread that you read before posting. Apologies for the confusion.

A little assistance please.

No instructions are included. I will guess that you are expected to solve for the value of the variable "x", and possibly also find the values of the various indicated angles.

drawing:

        B
        * 
       '  `
      '2x   `
     '        `
    '           `
   '              `
  '                 `
 '76                  `   5x - 2
*-----------------------*----------*
A                       C          D

Since C is supplementary I tried: 5x- 2 + x = 180.

I will guess that, by "C", you mean "angle BCD" rather than "angle ACB". But on what basis have you decided that the measure of angle ACB is "x"? All the triangle angle-sum rule gives you is that m(angle ACB) = 180 - 2x - 76, and all the supplementary angle-sum rule gives you is that (180 - 2x - 76) + (5x - 2) = 180. Neither of these seems to lead to your result...?

Instead, using the known-good rules, one gets:

. . . . .5x - 2 = 2x + 76

. . . . .3x = 78

...and so forth. Does this relate at all to whatever you got? Thank you!