A line that is tangent to each of the two coplanar circles
Square ABCD and square EFGH will reflect onto themselves across 8 lines of reflection
Answer:
I’m pretty sure the answer is A.
Hope this helps!
Let ABC be a triangle in the 3rd quadrant, right-angled at B.
So, AB-> Perpendicular BC -> Base AC -> Hypotenuse.
Given: sinθ=-3/5 cosecθ=-5/3
According to Pythagorean theorem, square of the hypotenuse is equal to the sum of square of the other two sides.
Therefore in triangle ABC, 〖AC〗^2=〖AB〗^2+〖BC〗^2 ------
--(1)
Since sinθ=Perpendicular/Hypotenuse ,
AC=5 and AB=3
Substituting these values in equation (1)
〖BC〗^2=〖AC〗^2-〖AB〗^2
〖BC〗^2=5^2-3^2
〖BC〗^2=25-9
〖BC〗^2=16
BC=4 units
Since the triangle is in the 3rd quadrant, all trigonometric ratios, except tan
and cot are negative.
So,cosθ=Base/Hypotenuse Cosθ=-4/5
secθ=Hypotnuse/Base secθ=-5/4
tanθ=Perpendicular/Base tanθ=3/4
cotθ=Base/Perpendicular cotθ=4/3
Let’s use process of elimination
Since the y intercept is at (0, y)
It is around -9 and -10
It cannot be A or C
Now for slope, it is moving up at a steep pace. The slope should be positive.
For D, the slope is too little.
Therefore, it has to be B
Solution: B
