SOLUTION: Use the drawing in which AC intersects DB at point O to answer the question. Two lines intersect at point O to form four angles. Line A C goes from point A in the top left,

Algebra ->  Angles -> SOLUTION: Use the drawing in which AC intersects DB at point O to answer the question. Two lines intersect at point O to form four angles. Line A C goes from point A in the top left,      Log On


   



Question 1191300: Use the drawing in which
AC intersects DB
at point O to answer the question.
Two lines intersect at point O to form four angles.
Line A C goes from point A in the top left, through point O in the center, to point C in the bottom right.
Line D B goes from point D in the bottom left, through O on line A C, to point B in the top right.
The angle at the bottom, ∠C O D, is labeled 1.
The angle on the left, ∠D O A, is labeled 2.
The angle at the top, ∠A O B, is labeled 3.
The angle on the right, ∠B O C, is labeled 4.
If
m∠2 =
x
2
− 30
°

and
m∠3 =
x
3
+ 60
°

,
find x and m∠2 in degrees.
x =
m∠2 =
°
Need Help? Read It

Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
Here's how to solve for x and m∠2:
1. **Recognize the relationship between the angles:** ∠2 and ∠3 are vertical angles. Vertical angles are congruent, meaning they have the same measure.
2. **Set up an equation:** Since m∠2 = m∠3, we can write:
x² - 30 = x³ + 60
3. **Rearrange the equation:** To solve for x, we need to rearrange the equation. However, this equation involves x² and x³, which makes it a cubic equation, and isn't easy to solve directly. The problem likely intended a linear relationship. It is possible there was a typo in the original problem. If we assume that the intended relationship was linear rather than polynomial, and assuming that the intended measures were m∠2 = x - 30 and m∠3 = x + 60, then we can proceed as follows:
x - 30 = x + 60
4. **Solve for x:**
Subtract x from both sides:
-30 = 60 This is not possible.
It's possible that there was a typo in the original problem. If the angle measures were instead m∠2 = x - 30 and m∠3 = 2x + 60, then we would proceed as follows:
x - 30 = 2x + 60
-90 = x
5. **Substitute x back into the equation for m∠2:**
m∠2 = x - 30
m∠2 = -90 - 30
m∠2 = -120°
Since angle measures can't be negative, it's likely there was a typo in the original problem. Double-check the given angle measures. If they were m∠2 = x-30 and m∠3 = x+60, then there is no solution. If they were m∠2 = x-30 and m∠3 = 2x+60, then x = -90 and m∠2 = -120.
**If you can provide the correct angle measures, I can help you solve the problem accurately.**

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

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