jugoslav

https://web2.0calc.com/questions/geometry_99280

https://web2.0calc.com/questions/help-is-needed_5

area = 13pi + pi/4 = 41.626 m²

Let ABC be a right triangle such that B is a right angle. A circle with a diameter of  BC meets side AC at D.  If the area of ABC is 150 and AC = 40 then what is BD?

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Area of Δ ABC:      A = 150 u²

The base of a triangle ABC:       AC = 40 units

BD = 2A / AC = 300 / 40          BD = 7.5 units  

(180/9 * 1) + (180/9 * 3) + (180/9 * 5) = 180º

Thanks!

You did something clever right here:> https://web2.0calc.com/questions/pre-cal-need-help

urw, Phill!

Important: When a tangent and a secant, two secants, or two tangents intersect outside a circle then the measure of the angle formed is one-half the positive difference of the measures of the intercepted arcs.

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∠SOT = xº         ∠ROB = (4x)º            ∠RTB = 28º

 ∠RTB = 1/2 (∠ROB - ∠SOT)     

28º = 1/2 (4x - x)            x = 18.6666666

Arc RS = 180 - 5x = 86.6666666º

AF = sinB * c

BF = AF / cosB 

CF = sqrt(b² - AF²)

BC = BF + CF

When you have all 3 sides, use the law of cosines to calculate the other 2 angles.

y / 20 = 4 / y

y² = 80

y = √80