tan(-60 degrees) = - sqrt(3)
1.732
- tan 60
Yes
5400
tan(pi/3) = tan (60 degrees) = 1.732 which is square root of 3
cot 115 deg = - tan25 deg
The exact value of 60 degrees would be 1/2. This is a math problem.
Other side = 60 tan 20, so area of rectangle = 3600 tan 20 = 3600 x 0.36397= 1310.29 sq ft
sin(60) or sin(PI/3) = sqrt(3)/2 cos(60) or cos(PI/3)=1/2 tan(60) or tan(PI/3) = sin(60)/cos(60)=sqrt(3) But we want tan for -sqrt(3). Tangent is negative in quadrant II and IV. In Quadrant IV, we compute 360-60=300 or 2PI-PI/3 =5PI/3 tan(5PI/3) = -sqrt(3) Tangent is also negative in the second quadrant, so we compute PI-PI/3=2PI/3 or 120 degrees. tan(t)=-sqrt(3) t=5PI/3 or 2PI/3 The period of tan is PI The general solution is t = 5PI/3+ n PI, where n is any integer t = 2PI/3+ n PI, where n is any integer
A pentagon has 5 sides. The perimeter is 60 so each of its sides is 60/5=12. Area = n (s/2)^2 / tan( π /n) = 5(12/2)^2 / tan ( π /5) = 247.7487
Sin(30) = 1/2 Sin(45) = root(2)/2 Sin(60) = root(3)/2 Cos(30) = root(3)/2 Cos(45) = root(2)/2 Cos(60) = 1/2 Tan(30) = root(3)/3 Tan(45) = 1 Tan(60) = root(3) Csc(30) = 2 Csc(45) = root(2) Csc(60) = 2root(3)/3 Sec(30) = 2root(3)/3 Sec(45) = root(2) Sec(60) = 2 Cot(30) = root(3) Cot(45) = 1 Cot(60) = root(3)/3