Both circles have the same center. The circumference of the inner circle is 25.12 millimeters.
1 answer:
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Answer:
Match each of the trigonometric expressions below with theequivalent non-trigonometric function from the following list.Enter the appropiate letter(A,B, C, D or E)in each blank
A . tan(arcsin(x/8))
B . cos (arsin (x/8))
C. (1/2)sin (2arcsin (x/8))
D . sin ( arctan (x/8))
E. cos (arctan (x/8))
These are the spaces to fill out :
.. ..........x/64 (sqrt(64-x^2))
.............x/sqrt(64+x^2)
.............sqrt(64-x^2)/8
..............x/sqrt(64-x^2)
..............8/sqrt(64+x^2)
A. ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)
B . cos (arsin (x/8)) ....sqrt(64-x^2)/8
Step-by-step explanation:
To solve this we have to find the missing sides to each of the triange discribed in prenthesis thus
A we have the sides of the triangle given by x, 8 and or
thus tan(arcsin(x/8)) = =
Therefore ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)
B
Here we have cos = adjacent/hypotenuse where adjacent side is and hypothenuse = 8 we have
/8
B . cos (arsin (x/8)) ....sqrt(64-x^2)/8
