For the figures that have rotational symmetry, list the angles of rotation less than 360 degrees. For figures without rotational
symmetry, write “no rotational symmetry”.
Please show all work and not just the answers.
Please show all work and not just the answers.
1 answer:
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CE ⊥ CF because m∠ECA + m∠FCA = 90°
Step-by-step explanation:
In ∆ABC
- On the extension of the side BC , draw a line segment CD ≅ CA
- Draw the segment AD
- The line segment CE is the angle bisector of ∠ACB
- The line segment CF is the median towards AD in ∆ ACD
We want to prove that CF ⊥ CE
Look to the attached figure
In Δ ABC
∵ CE is the bisector of angle ACB
∴ ∠ACE ≅ ∠BCE
In Δ ACD
∵ CA = CD
∴ Δ ACD is an isosceles triangle
∵ AD is the median towards AD
- In any isosceles triangle the median from a vertex to its opposite
side bisects this vertex
∴ AD bisects ∠ACD
∴ ∠ACF ≅ ∠DCF
∵ BCD is a straight segment
∵ CE , CA , CF are drawn from point C
∴ m∠BCE + m∠ACE + m∠ACF + m∠DCF = 180°
∵ m∠ACE ≅ m∠BCE
∵ m∠ACF ≅ m∠DCF
- Replace m∠BCE by m∠ACE and m∠DCF by m∠ACF
∴ m∠ACE + m∠ACE + m∠ACF + m∠ACF = 180°
∴ 2 m∠ACE + 2 m∠ACF = 180°
- Divide all terms by 2
∴ m∠ACE + m∠ACF = 90°
∴ EC ⊥ CF
CE ⊥ CF because m∠ECA + m∠FCA = 90°
Learn more:
You can learn more about perpendicular lines in brainly.com/question/11223427
#LearnwithBrainly
Answer:
61.08m
Step-by-step explanation:
Let " l " be the length of the rope and " θ " the angle at which it is raised. In other words:
- l = 65m
- θ = 70°
As you can see, there has been made a right angle triangle. So:
So the kite is approximately 61.08m above the ground.
In order to find the sin70° i used a calculator.
