The sine of an angle is equal to the cosine of the complementary angle.
That is; sin r = cos 90 -r , where r is an acute angle
Therefore;
sin 73 = cos (90-73)
sin 73 = cos 17
Hence the value of r = 17
Answer:
m∠1 = 25° (opposite angles are congruent)
m∠2 = 87° ⇒ 180 - (25 + 68)
m∠3 = 68° ⇒ 180 - (87 + 25)
Hope this helps!
Given:
An angle θ on the unit circle has a sine of
.
To find:
The value of cosθ.
Solution:
We have,
![\sin \theta=\dfrac{1}{4}](https://tex.z-dn.net/?f=%5Csin%20%5Ctheta%3D%5Cdfrac%7B1%7D%7B4%7D)
Since sinθ is positive, therefore, θ lies in first or second quadrant.
We know that,
![\sin^2 \theta +\cos^2 \theta =1](https://tex.z-dn.net/?f=%5Csin%5E2%20%5Ctheta%20%2B%5Ccos%5E2%20%5Ctheta%20%3D1)
![(\dfrac{1}{4})^2+\cos^2 \theta =1](https://tex.z-dn.net/?f=%28%5Cdfrac%7B1%7D%7B4%7D%29%5E2%2B%5Ccos%5E2%20%5Ctheta%20%3D1)
![\dfrac{1}{16}+\cos^2 \theta =1](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B16%7D%2B%5Ccos%5E2%20%5Ctheta%20%3D1)
![\cos^2 \theta =1-\dfrac{1}{16}](https://tex.z-dn.net/?f=%5Ccos%5E2%20%5Ctheta%20%3D1-%5Cdfrac%7B1%7D%7B16%7D)
Taking square root on both sides.
![\cos \theta =\pm \sqrt{\dfrac{16-1}{16}}](https://tex.z-dn.net/?f=%5Ccos%20%5Ctheta%20%3D%5Cpm%20%5Csqrt%7B%5Cdfrac%7B16-1%7D%7B16%7D%7D)
![\cos \theta =\pm \dfrac{\sqrt{15}}{4}](https://tex.z-dn.net/?f=%5Ccos%20%5Ctheta%20%3D%5Cpm%20%5Cdfrac%7B%5Csqrt%7B15%7D%7D%7B4%7D)
Therefore, the value of cosθ is either
or
.
Answer:
1:3
Step-by-step explanation:
It’s ratios.
A: 3
B: 9
36
129
C: x3
If this helps at all, the problem can equal out to 0 if x=7, hope this helps. :)