Which of the following can be inserted to make the statement true
$$\arccos \left( \frac { \sqrt { 36 - x ^ { 2 } } } { 6 } \right) = \arcsin ( \_\_\_\_\_ ) , 0 \leq x \leq 6$$
A) $\frac { x ^ { 2 } } { 6 }$
B) $\frac { x } { 6 }$
C) $36 - x ^ { 2 }$
D) $\frac { \sqrt { 72 - x ^ { 2 } } } { x }$
E)x

Question

Which of the following can be inserted to make the statement true
 $$\arccos \left( \frac { \sqrt { 36 - x ^ { 2 } } } { 6 } \right) = \arcsin ( \_\_\_\_\_ ) , 0 \leq x \leq 6$$
A) $\frac { x ^ { 2 } } { 6 }$
B) $\frac { x } { 6 }$
C) $36 - x ^ { 2 }$
D) $\frac { \sqrt { 72 - x ^ { 2 } } } { x }$
E)x

Quizplus · Accepted Answer

The correct choice is B) $\frac { x } { 6 }$. This is because $\arccos \left( \frac { \sqrt { 36 - x ^ { 2 } } } { 6 } \right)$ and $\arcsin \left( \frac { x } { 6 } \right)$ are complementary angles in a right triangle where $\frac { \sqrt { 36 - x ^ { 2 } } } { 6 }$ represents the adjacent side over the hypotenuse, and $\frac { x } { 6 }$ represents the opposite side over the hypotenuse. Since $\sin(\theta) = \cos(90^\circ - \theta)$, these expressions are equivalent in this context.

Which of the Following Can Be Inserted to Make the Statement