Answer:
Option A. 
Step-by-step explanation:
we know that
The lateral area of a cone is equal to

we have


substitute and solve for l (slant height)

simplify


Which question in order for me 2 answer
Answer:
Step-by-step explanation:
9) PQR Is an isosceles triangle
=> ∠PRQ = (180° - x)/2
PRS is an isosceles right triangle
=> ∠PRS = 45°
Have: ∠PRS + ∠PRQ = 115°
=> 
=> 180 - x = (115 - 45).2 = 140
<=> x = 180 - 140 = 40
10) ABD is an isosceles right triangle => ∠ABD = 45°
BCD is an equilateral triangle => ∠CBD = 60°
have: x = ∠ABD + ∠CBD = 45° + 60° = 105°
11) have: x = y (2)
PQT is an isosceles triangle => ∠PQT = 180 - 70.2 = 40
QTS is an isosceles triangle => ∠TQS = 180 -2x
QRS is an isosceles triangle => ∠RSQ = y
have: 40 + 180 - 2x + y = 180 => 2x - y = 40 (1)
(1)(2) => 
=> x + y = 80
12) EFJ Is an equilateral triangle => ∠FJE = 60
∠FJE is the outer angle of the triangle FHJ but FHJ is an isosceles triangle
=> 60 = 2.∠JHF => ∠JHF = 30°
∠JHF is the outer angle of the triangle FHG
=> 30° = 2x
<=> x = 15°
Answer:
Reference angles are always acute.
We can find the reference angle with the "formula"
π
−
θ
, where
θ
is our given angle.
We can plug
5
π
6
into our expression to get
π
−
5
π
6
=
6
π
6
−
5
π
6
=
π
6
Thus, our reference angle is
π
6
15 3/4% = 15.75%
15.75% = 0.1575
The answer is A.