a = 180-12-2-46-8
a = 180-68
a = 112°
a is angle ABC.
Let AC be 1 unit.
It implies;
(1/sin112) = (b/sin14)
b = 0.257447904 units.
b is AB.
c = 180-8-2
c = 170°
c is angle ADC.
It implies;
(1/sin...
Online Edumath Educators and Learners are Super Smart and Amazingly, Very Clever.
Communicate us to mentor/teach your child/children Mathematics online at affordable tuition, helping them become a l...
tana = 12/8
a = atan(3/2)°
b = 180-a
b = (180-atan(3/2))°
(10/sin(180-atan(3/2))) = (4/sinc)
c = 19.4400348282°
d = 180-b-c
d = 180-19.4400348282-(180-atan(3/2))
d = 36.8698976458°
e = 90-d
e =...
Shaded area is;
Area under the curve at points (-1, 0) and (7, 0) - Area triangle with height 4 unit and base 8 unit.
= ⅓(208) - ½(4*8)
= ⅓(160) space unit.
Area S is;
Area under the curve at p...
Let the side of the regular hexagon be 1 unit.
Therefore;
Area blue is;
2(area triangle with height 1 unit and base (⅓√(3)) unit).
= 2(½*1*(⅓√(3)))
= ⅓√(3) space unit.
Area orange is;...
Area blue is:
Area triangle with height 1 cm and base (3sin(109.471220634)) cm.
= 0.5*1*3sin(109.471220634)
= 1.41421356238 cm²
= √(2) cm²
Let the radius of the inscribed white semi circle be r.
Therefore;
r²+(1+r)²=(2+r)²
r²+1+2r+r²=4+4r+r²
r²-2r-3 = 0
(r-1)² = 3+(-1)²
(r-1)² = 4
r = 1±2
r ≠ -1
r = 3 unit.
It implies;
Area blue is;...
Radius of the inscribed circle, r is;
2r² = 4
r² = 2
r = √(2) unit.
Therefore;
Area R in square unit to 3 d. p. is;
Area sector with radius √(2) unit and angle atan(0.5√(2))° - Area triangle with...
tan60 = a/10
a = 10√(3) cm.
a is the height of the equilateral triangle.
sin(atan(2)) = b/(10√(3))
b = 15.4919333848 cm.
b is a side length of the square.
It implies;
Area Square is;
b...
Let the base of ABCD be 1 unit.
a = 180-24-38-48
a = 180-110
a = 70°
b = 38+48
b = 86°
(1/sin70) = (c/sin86)
c = 1.0615854889 units.
d = 180-26-24-38
d = 180-88
d = 92°
(1/sin92) = (e/sin38)
e...
