Trig solve these triangles any or all

  see below

Step-by-step explanation:

21) The law of sines can be used, since you have a side and its opposite angle.

  sin(F)/DE = sin(D)/EF

  F = arcsin(DE/EF·sin(D)) = arcsin(20/31·sin(95°)) ≈ 39.994°

  E = 180° -95° -39.994° ≈ 45.006°

  DF = sin(45.006°)/sin(95°)·31 ≈ 22.006

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22) The remaining two problems can be solved using the law of cosines:

  c^2 = a^2 + b^2 - 2ab·cos(C)

Of course, c is the square root of the expression on the right.

  EF = √(19^2 +35^2 -2(19)(35)cos(61°)) ≈ √(941.203) ≈ 30.679

Then an angle can be found using the law of sines

  E ≈ arcsin(35/30.679·sin(61°)) ≈ 86.203°

  F ≈ 180° -61° -86.203° ≈ 32.797°

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23) As in 22 …

  RS = √(20^2 +28^2 -2(20)(28)cos(91°)) ≈ √(1203.547) ≈ 34.692

  R ≈ arcsin(20/34.692·sin(91°)) ≈ 35.199°

  S ≈ 180° -91° -35.199° ≈ 53.801°

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.

step-by-step explanation:   divide the numerator by the denominator to get the decimal

it would still be 193

if divide by 100 then 1.93

i am not sure