If StartFraction pi Over 2 EndFraction less than t less than pi , which of the following is true?
1 answer:
Answer:
b. cosine t less than 0 and cotangent t greater than 0
Step-by-step explanation:
We have the following relation
if we apply the cosine function in the relation we get:
the cosine of t is between 0 and -1 then (cosine t less than 0)
If we now apply cotangent function in the relation:
This means that cotang is greater than 0, therefore the correct answer is b. cosine t less than 0 and cotangent t greater than 0
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Answer:
Hence proved △ABE∼△CBF.
Step-by-step explanation:
Given,
ABCD is a parallelogram.
BF ⊥ CD and
BE ⊥ AD
To Prove : △ABE∼△CBF
We have drawn the diagram for your reference.
Proof:
Since ABCD is a parallelogram,
So according to the property of parallelogram opposite angles are equal in measure.
⇒1
And given that BF ⊥ CD and BE ⊥ AD.
So we can say that;
⇒2
Now In △ABE and △CBF
∠A = ∠C (from 1)
∠E = ∠F (from 2)
So by A.A. similarity postulate;
△ABE∼△CBF
