Hint: $\bf \begin{array}{llll}
sin^{-1}[sin(whatever)]= whatever\\\\
cos^{-1}[cos(whatever)]= whatever\\\\
tan^{-1}[tan(whatever)]= whatever
\end{array}\\\\
------------------\\\\
\begin{array}{llll}
sin[sin^{-1}(whatever)]= whatever\\\\
cos[cos^{-1}(whatever)]= whatever\\\\
tan[tan^{-1}(whatever)]= whatever
\end{array}$

3 0

2 years ago

Identify the area of segment MNO to the nearest hundredth. HELP PLEASE!! I don't understand it!

katrin2010 [14]

Answer:

13.98 in²

Step-by-step explanation:

I don't understand it, either.

Point N is part of a "segment" that above and to the right of chord MO. It is the sum of the areas of 3/4 of the circle and a right triangle with 7-inch sides. The larger segment MO to the upper right of chord MO has an area of about 139.95 in², which is not an answer choice.

The remaining segment, to the lower left of chord MO does not seem to have anything to do with point N. However, its area is 13.98 in², which is an answer choice. Therefore, we think the question is about this segment, and we wonder why it is called MNO.

The area of a segment is given by the formula ...

A = (1/2)(θ -sin(θ))r² . . . . . . where θ is the central angle in radians.

Here, we have θ = π/2, r = 7 in, so we can compute the area of the smaller segment MO as ...

A = (1/2)(π/2 -sin(π/2))(7 in)² = 24.5(π/2 -1) in² ≈ 13.9845 in²

Rounded to hundredths, this is ...

≈ 13.98 in²

3 0

2 years ago