Your answer is B.
The area of the shaded region will be the area of the circle minus the area of the triangle inside the circle. Then:
The circle has radius 4 cm (distance from the center to the edge of the circle), so the area of the circle is
Now, the area of the triangle is
The base of the triangle is the diameter of the circle (as you can see in the image) and the height is the radius of the circle. Then the are of the triangle is
Finally, the shaded area is
=
.
Answer:
$16.80
Step-by-step explanation:
30% of 24 and the answer is subtracted from $24
Answer:
Height of tree is
<em>15 m.</em>
<em></em>
Step-by-step explanation:
Given that student is 20 m away from the foot of tree.
and table is 1.5 m above the ground.
The angle of elevation is: 34°28'
Please refer to the attached image. The given situation can be mapped to a right angled triangle as shown in the image.
AB = CP = 20 m
CA = PB = 1.5 m
= 34°28' = 34.46°
To find TB = ?
we can use trigonometric function tangent to find TP in right angled ![\triangle TPC](https://tex.z-dn.net/?f=%5Ctriangle%20TPC)
![tan \theta = \dfrac{Perpendicular}{Base}\\tan C= \dfrac{PT}{PC}\\\Rightarrow tan 34.46^\circ = \dfrac{PT}{20}\\\Rightarrow PT = 20 \times 0.686 \\\Rightarrow PT = 13.72\ m](https://tex.z-dn.net/?f=tan%20%5Ctheta%20%3D%20%5Cdfrac%7BPerpendicular%7D%7BBase%7D%5C%5Ctan%20C%3D%20%5Cdfrac%7BPT%7D%7BPC%7D%5C%5C%5CRightarrow%20tan%2034.46%5E%5Ccirc%20%3D%20%5Cdfrac%7BPT%7D%7B20%7D%5C%5C%5CRightarrow%20PT%20%3D%2020%20%5Ctimes%200.686%20%5C%5C%5CRightarrow%20PT%20%3D%2013.72%5C%20m)
Now, adding PB to TP will give us the height of tree, TB
Now, height of tree TB = TP + PB
TB = 13.72 + 1.5 = 15.22
<em>15 m</em>
Answer:
When rotating a point 90 degrees counterclockwise about the origin our point A(x,y) becomes A'(-y,x). In other words, switch x and y and make y negative
Step-by-step explanation:
We know that
cos a+cos b=cos[(a+b)/2]*cos[(a-b)/2]
we have
<span>cos(π/7)+cos(2π/7)+cos(3π/7)+cos(4π/7)-------------> equation 1
</span>cos(4π/7)+cos(2π/7)=cos[(4π/7+2π/7)/2]*cos[(4π/7-2π/7)/2]
=cos(3π/7)*cos(π/7)
then
cos(4π/7)+cos(2π/7)=cos(3π/7)*cos(π/7)--------------> equation 2
[cos(3π/7)+cos(π/7)]=cos[(3π/7+π/7)/2]*cos[(3π/7-π/7)/2]
=cos(2π/7)*cos(π/7)
then
[cos(3π/7)+cos(π/7)]=cos(2π/7)*cos(π/7)-----------> equation 3
I substitute 2 and 3 in 1
[cos(3π/7)+cos(π/7)]+[cos(4π/7)+cos(2π/7)]
{cos(2π/7)*cos(π/7}+{cos(3π/7)*cos(π/7)}
=cos(π/7)*[cos(2π/7)+cos(3π/7)]
the answer is
cos(π/7)+cos(2π/7)+cos(3π/7)+cos(4π/7)=cos(π/7)*[cos(2π/7)+cos(3π/7)]