Answer: sin
= ± 
Step-by-step explanation:
We very well know that,
cos2A=1−2sin²A
⟹ sinA = ±
As required, set A = & cos a= 
,thus we get
sin =± 
∴ sin =±
= ±
since ,360° < <450°
,180° < <225°
Now, we are to select the value with the correct sign. It's is obvious from the above constraints that the angle a/2 lies in the III-quadrant where 'sine' has negative value, thus the required value is negative.
hope it helped!
The equation of the line would be
y = mx+b
where m is the slope, b is the y intercept.
<span>The line will form a triangular region in the first quadrant. Its area would be 1/2 base times height. The height is the y intercept
and the base is y intercept divided by slope. Therefore,</span>
A = b^2/2m
At point (2,5)
5 = 2m+b
Substitute that in the area
A = b^2/5-b
to find the least area, differentiate the area with respect to the height and equate it to 0
dA/db = 0
<span>find b and
use that to find m. Then, you can have the equation of the line.</span>
Volume = Length x Width x Height
Length = 4 ft
Width = 4 ft
Height = 15 ft
Volume = 4 ft x 4 ft x 15 ft = 240 ft^3
