jugoslav

In right triangle DEF, we have angle E = 90 degrees and tan D = 4*sin D. What is sin F?

Angle F = 90º - [cos^-1(1/4)]  ==>  ∠F = 14.47751219º

Points A(0,0), B(9,6) and C(6,12) are vertices of triangle ABC. Point D is on segment AB such that 3(AD) = DB, point E is on segment BC such that 3(BE) = EC, and point F is on segment CA such that 3(CF) = FA. What is the ratio of the area of triangle DEF to the area of triangle ABC? Express your answer as a common fraction.

Area of ΔABC = 36 u²             area of ΔDEF = 15.75 u²

So, the ratio...         [DEF] / [ABC] = 7/16 

Good job, Guest!

Use the law of sines to find AC, then use the same law to find the length of side AB.

170º * 2 = 340º    This is a major arc.

Minor arc is         360º - 340º = 20º

Minor arc {AB} = (36pi/360)*20 = 2pi

btw, major arc = 34pi

Large circle radius R = ?        Small circle radius r = 1

Large circle area            A_l = R²pi           Small circle area      A_s = pi

Trapezoid area               A_t = R²/2 + R + 1/2

Annular ring area           A_r = R²pi - pi

R²pi - pi  / (R²/2 + R + 1/2) = 3/2

R = 1.62719711 units

So, the ratio is      1 : 1.62719711       or        0.614553697

 The ratio of the volume of the new polyhedron to the volume of the original tetrahedron is:

6.425068997 : 7.54247232       or       0.851851851        or         851 / 999

An equilateral triangle of side length 6 is rotated about an altitude to form a cone. What is the volume of the cone?

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Cone radius         r = 3 

Slant height         h_s = 6               ( Height  h = √27 )

Cone volume         V = 1/3 * pi * r² * sqrt(h_s² - r²)    ==>  V = 48.97258284 u³

                                         or

                              V = r² * pi * h / 3        ===> V = 48.97258284 u³

Let ABCDEFGH be a cube of side length 5, as shown. Let P and Q be points on line AB and line AE, respectively, such that AP = 3 and AQ = 1. The plane through C, P, and Q intersects line DH at R. Find DR.

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∠CPQ ≈ 68.199º

∠DCR ≅ ∠APQ = 18.43494881º

DR = tan(∠DCR) * CD = 1.66666667    or     5/3

CORRECTIONS:

The perimeter of triangle A₃B₃C₃ is 14 times the perimeter of triangle ABC, and the triangle is turned upside-down.

P = 224 units

A (0, 0)      B (3, 4)        C (6, 0) 

A₁ (5.76, - 7.68)        B₁ (4, - 3)         C₁ (0, - 6)   

A₂ (13.76, - 7.68)           B₂ (12, - 3)          C₂ (8, - 6)   

A₃ (-192.64, 107.52)           B₃ (-168, 42)            C₃ (-112, 84)

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