HELP ME ASAP.. MY TEACHER IS ASKING FOR THE ANSWER

a) 10.7 cm

b) 11.6 cm

c) 29.9 cm²

Step-by-step explanation:

a) sin(α)/a = sin(β)/b

sin(∠BCD)/BD = sin(DBC)/DC

sin(41°)/BD = sin(55°)/13.4

BD = 13.4*sin(41°)/sin(55°) = 10.73 cm ≈ 10.7 cm

b) From ΔBCD , ∠BDC = 180-(41+55) = 84°

∠ABD and ∠BDC are alternate interior angles, so they are congruent, and

∠ABD = ∠BDC  = 84°

AD²= AB² + BD² - 2*AB*BD*cos (∠ABD) =

= 5.6² + 10.73² - 2*5.6*10.73*cos(84°) = 133.93 cm²

AD =√(133.93) ≈11.6 cm

c) Area(ADB) = (1/2)*AB*BD*sin(∠ABD)=(1/2)*5.6*10.73*sin(84°) ≈ 29.9 cm²

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step-by-step explanation:

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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.