I don't get how it would be a system of equations.
Answer:
m∠ABE = 27°
Step-by-step explanation:
* Lets look to the figure to solve the problem
- AC is a line
- Ray BF intersects the line AC at B
- Ray BF ⊥ line AC
∴ ∠ABF and ∠CBF are right angles
∴ m∠ABF = m∠CBF = 90°
- Rays BE and BD intersect the line AC at B
∵ m∠ABE = m∠DBE ⇒ have same symbol on the figure
∴ BE is the bisector of angle ABD
∵ m∠EBF = 117°
∵ m∠EBF = m∠ABE + m∠ABF
∵ m∠ABF = 90°
∴ 117° = m∠ABE + 90°
- Subtract 90 from both sides
∴ m∠ABE = 27°
Answer:
1. True, 2. False, 3. I am blind.
Step-by-step explanation:
For the 3rd one I cannot confirm becaue I am blind.
Answer:
∠1 is 33°
∠2 is 57°
∠3 is 57°
∠4 is 33°
Step-by-step explanation:
First off, we already know that ∠2 is 57° because of alternate interior angles.
Second, it's important to know that rhombus' diagonals bisect each other; meaning they form 90° angles in the intersection. Another cool thing is that the diagonals bisect the existing angles in the rhombus. Therefore, 57° is just half of something.
Then, you basically just do some other pain-in-the-butt things after.
Since that ∠2 is just the bisected half from one existing angle, that means that ∠3 is just the other half; meaning that ∠3 is 57°, as well.
Next is to just find the missing angle ∠1. Since we already know ∠3 is 57°, we can just add that to the 90° that the diagonals formed at the intersection.
57° + 90° = 147°
180° - 147° = 33°
∠1 is 33°
Finally, since that ∠4 is just an alternate interior angle of ∠1, ∠4 is 33°, too.
