square root of 3
tan(-60 degrees) = - sqrt(3)
1.732
- tan 60
Yes
5400
Tan(60) = Sin(60)/ Cos(60) Sin(60) = sqrt(3)/2 Cos(60) = 1/2 Hence Sin(60) / Cos(60) = [sqrt(3) / 2] / [1/2} => sqrt(3) / 2 X 2/1 sqrt(3) Hence Tan(60) = sqrt(3) = Numerically = 1.732050808....
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