Describe a sequence of transformations that takes triangle ABC to triangle A'B'C
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Answer:
None of the Above
Step-by-step explanation:
As it is given that
cosФ= 3/2
and Ф lies in quadrant 4
Now in quadrant 4 we know that by the rules of trigonometry that
cos Ф is positive in 4th quadrant also in 4th quadrant
and sinФ is negative in 4th quadrant
also we know that by simple rules of trigonometry in a right angled triangle
cos Ф= =
now from the law of trigonometry
cos²Ф+sin²Ф=1
From this we derive the value of sinФ
solving the equation
sin²Ф=1-cos²Ф
Putting in the values of cos Ф
sin²Ф=1-
sin²Ф=1-
solving the fraction
sin²Ф=
sin²Ф=
so taking square root of both sides
here
=Ф
so it becomes
sin Ф=
as we know that in imaginary numbers
= ι
so the given becomes
sin Ф=
Which is the value of sin Ф
And in the given values we can see that it is none of the values
Also seeing the question we can see that the question is absurd because
cos Ф= base / hypotenuse
in our question we can see that base =3 and hypotenuse = 2
but in real geometry this is the rule that hypotenuse can never be smaller then base either it is equal to base or greater then base
the formula for finding value of finding hypotenuse is
hypotenuse ² = base ² + perp ²
so this formula shows that if perp is 0 then hypotenuse will be equal to base
in other cases it would be greater then base