Find the exact value of each trigonometric function

Mathematics

1 answer:

GaryK [48]3 years ago

3 0

Answer:

excuse me what

Step-by-step explanation:

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Suppose a square ABCD has a side length of 1 unit. The arcs BD and AC are circular arcs with centers at A and D, respectively. F

Likurg_2 [28]

Answer:

$1-\frac{\sqrt{3} }{4}-\frac{\pi}{6}$

Step-by-step explanation:

The missing figure is shown in the attachment

The area of the shaded region = Area of Square - (Area of sector AOB + Area of equilateral triangle BOC + Area of sector COD)

Area of Sector AOB=Area of Sector COD= $\frac{30}{360}*\pi*1^2=\frac{\pi}{12}$

Area of equilateral triangle = $\frac{1}{2}*r*\frac{\sqrt{3} }{2}*r= \frac{1}{2}*1*\frac{\sqrt{3} }{2}*=\frac{\sqrt{3} }{4}$

Area of shade region = $1^2-\frac{\sqrt{3} }{4}-\frac{\pi}{12}*2$

$1-\frac{\sqrt{3} }{4}-\frac{\pi}{6}$

6 0

3 years ago

Find the distance between<br> 2-3i<br> And<br> 9+21i

Goryan [66]

2 - 3i is just another of writing (2, -3) in the cartesian plane, the 2-3i however is using the "imaginary axis" for the imaginary plane, is all however the imaginary axis is simply the equivalent of the y-axis. Likewise 9+21i is just (9,21), so let's just use the distance formula.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{9}~,~\stackrel{y_2}{21})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[9-2]^2+[21-(-3)]^2}\implies d=\sqrt{(9-2)^2+(21+3)^2} \\\\\\ d=\sqrt{7^2+24^2}\implies d=\sqrt{49+576}\implies d=\sqrt{625}\implies d=25$