For -7pi/6 is an angle in second quadrant, then sine and cosecant must be positive; and cosine, secant, tangent and cotangent must me negative.
The reference angle is:
7pi/6-pi=7pi/6-6pi/6=(7pi-6pi)/6=pi/6
Then
sin(-7pi/6)=sin(pi/6)→sin(-7pi/6)=1/2
cos(-7pi/6)=-cos(pi/6)→cos(-7pi/6)=-sqrt(3)/2
csc(-7pi/6)=1/sin(-7pi/6)=1/(1/2)=1(2/1)=2/1→csc(-7pi/6)=2
sec(-7pi/6)=1/cos(-7pi/6)=1/(-sqrt(3)/2)=-1(2/sqrt(3))=-2/sqrt(3)→
sec(-7pi/6)=-[2/sqrt(3)]*sqrt(3)/sqrt(3)=-2sqrt(3)/[sqrt(3)]^2→
sec(-7pi/6)=-2sqrt(3)/3
tan(-7pi/6)=sin(-7pi/6)/cos(-7pi/6)=(1/2)/(-sqrt(3)/2)=-(1/2)*(2/sqrt(3))→
tan(-7pi/6)=-2/[2sqrt(3)]=-1/sqrt(3)=-[1/sqrt(3)]*[sqrt(3)/sqrt(3)]→
tan(-7pi/6)=-sqrt(3)/[sqrt(3)]^2→tan(-7pi/6)=-sqrt(3)/3
cot(-7pi/6)=cos(-7pi/6)/sin(-7pi/6)=[-sqrt(3)/2]/(1/2)=-sqrt(3)/2*(2/1)→
cot(-7pi/6)=-2sqrt(3)/2→cot(-7pi/6)=-sqrt(3)
Answers:
sin(-7pi/6) = 1/2
cos(-7pi/6) = - sqrt(3)/2
tan(-7pi/6) = - sqrt(3)/3
csc(-7pi/6) = 2
sec(-7pi/6) = - 2*sqrt(3)/2
cot(-7pi/6) = - sqrt(3)
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Answer:
x = 16
Step-by-step explanation:
The angle bisector divides the sides proportionally.
(x -4)/9 = x/12
4(x -4) = 3x . . . . . . . multiply by 36
x = 16 . . . . . . . . . . . . add 16-3x to both sides, simplify
Okay so you do not really need the circle equation. if you make a triangle with it's x length as 3 and it's y length as 4, you will be able to find the third length. Do Pythagorean theorem to find the hypotenuse.
the hypotenuse will be five after you calculate it.
sin is opposite over hypotenuse or Y over R
so... the sin is 4/5
make sure you know what quadrant your triangle is in for the negatives
A right triangles hypotenuse is the square root of both of the other side squared. The formula for this is a^2 + b^2= c^2, with c being the longest side. So 10 squared is 100, 13 squared is 169, but. 17 squared is 289. The sum of the other two sides are 269. So it doesn’t equal
