jugoslav

Given: ΔАВС, m∠ACB = 90° 
CD ⊥ AB, m∠ACD=60°,BC = 6 cm

 Find CD, Area of ΔABC 

CD = sin(60º) * 6

AC = tan(60º) * 6

[ABC] = (AC * AB) / 2

angle A = 30 degrees

In triangle ABC, angle ACB is 50 degrees, and angle CBA is 75 degrees. Let D be the foot of the perpendicular from A  to BC, O the center of the circle circumscribed about triangle ABC, and E the other end of the diameter which goes through A. Find the angle DAE, in degrees.

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∠ABC = 75º          ∠ACB = 50º           ∠BAC = 55º          Let AB be 8 units

1/   By using the law of sines we find that BC = 8.555 and AC = 10.088

2/   Using Heron's formula we calculate the area of [ABC] = 30.055 u²

3/   Inradius    r = A / s  ==>  r = 2.481           ( s -- semiperimeter)

4/   Circumradius      R = (abc) / 4rs     ==>    R = 5.222

∠DAC = 40º          ∠BAD = 15º

Angle OAC = cos-1[(AC/2) / AO]    ==>  ∠OAC = 15º

And finally...        ∠DAE = 55 - (2*15) = 25º

Find the equations of the median from vertex A to the opposite side BC using the points below:

A(9, 5), B(2, 5), C(4, 1)

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1/   Equation of the median   y = 1/3x + 2

2/   Length of the median (round your answer to the nearest tenth) = √40 = 6.3

Rhombus ABCD has a perimeter of 148, and one of its diagonals has a length of 20. How long is the other diagonal?

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d = 2{ sqrt [(148/4)²- (20/2)²]}

Angle AOB = 4 * 28 = 112 degrees

∠BAC ≅ ∠BCE = 48º

∠ABC  ≅ ∠ACB  ≅ ∠ACD = 66º

8 * 10 = 80 u²

(2 * 8) / 2 = 8 u²

Using Heron's formula you can find the area of ΔABC.    [ABC] = 11.314 u²

So far we know that the area of the (whole) figure is 3/2 pi units²

AB = cos(42º) * AD