ΔAOB is a right angled triangle. Therefore the Pythagorean Theorem applies in this situation.
θ is the angle from a standard position of the line OA
The length of the y component is √(1-0)2 +(-3-(-3))2] =√(12+ 02) = 1 A(-3,1) to B(-3,0) which is opposite
Then the length of the x-component is √[(-3-0)2 +(0-0)2] = √(9+0)= 3 B(-3,0) to O(0,0) which is adjacent
The length of vector OA is √[(-3-0)2 + (1-0)2] = √(9+1) = √(10) A(-3,1) to O(0,0) which is the hypotenuse of the triangle
θ = 180 - α
sinθ = sin(180-α) = opposite/hypotenuse = 1/√10
cosθ = adjacent/hypotenuse = -3/√10
tanθ = opposite/adjacent = 1/-3 = -1/3
α= arcsin(1/√10) ≈ 18
θ =180 -18 ≈162
Given:
The radius of the base of the given cone = 5 cm
The slant height of the cone = 19 cm
To find the surface area of the given cone.
Formula
The surface area pf the given cone is,
where, r be the radius and
l be the slant height.
Now,
Taking, r = 5, l = 19 and π = 3.14 we get,
sq cm
or, sq cm
or, sq cm
Rounding off to the nearest square centimeter, we get;
Hence,
The surface area of the cone is 377 sq cm.
Thats means angle BEC is 63. By the vertical angle theorem AED is the same. 3x-27 when x=30 is also 63.
The number that lies between that is 7
