Answer: c
Step-by-step explanation:
they are similar but to be congruent they need to be the same and therefore can not touch like that
Answer:
Step-by-step explanation:
To prove Δ ABC similar to ΔDBE we can consider
Segments AC and DE are parallel.
⇒ DE intersects AB and BC in same ratio.
AB is a transversal line passing AC and DE.
⇒∠BAC=∠BDE [corresponding angles]
Angle B is congruent to itself due to the reflexive property.
All of them are telling a relation of parts of ΔABC to ΔDBE.
The only option which is not used to prove that ΔABC is similar to ΔDBE is the first option ,"The sum of angles A and B are supplementary to angle C".
Answer:
1;infroei[q
Step-by-step explanation:
[qioerf1m['
Answer:
(- 1/2 , √3/2)
Step-by-step explanation:
t=20π/3
t' = 20π/3 - 6π = 2π/3 (120°) ... 2nd quadrant
If we start from (1,0) of unit circle, the coordinate of terminal point (x,y)
OF/OE = - cos 60° = x / 1 = - 1/2
FE/OE = sin 60° = y/1 = √3/2
A = 1
4/17=2/8.5
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