Dragan

Using the law of cosines we can find angles B and C.

∠B = 104.478°             ∠OBL = ∠B /2 = 52.239°

∠C = 46.567°               ∠OCL = ∠C /2 = 23.2835°

Using the law of sines we can find the side CO of the triangle COB.

CO = 3.266

Radius   r = sin(OCL) * CO = 1.291

NQ = MR = 2r = 2.582

CQ = NQ / tan(∠C) = 2.444

BQ = 4 - CQ = 1.556

Angle MBR = 180 - ∠B = 75.522°

BR = MR / tan(MBR) = 0.667

And finally...       MN = BQ + BR = 2.223      

never mind 

I misread the text !!!  

Phill's answer's correct!!! 

MX = (8*6)/10 = 4.8

MQ = sqrt(QX² - MX²) = 3.6          MQ = NL

NX = 2*MX + 10 = 19.6

LY = PQ - MQ = 6.4

NY = LY - MQ = 2.8

XY^2 = NX^2 + NY^2 = 392   

Hint: use an equilateral triangle with a side length of 7 

I started off with a 130° angle. 

 ∠B = 95°          ∠D = 138°          95+138 = 100%

 ∠B = 40.77253219%      and       ∠D = 59.22746781%

The ratio of angles =>      B : D =  ∠BCO : ∠DCO

Therefore:  ∠BCO = 53.00429185°     and     ∠DCO = 130° - ∠BCO = 76.99570815°

∠X = 78.99570815º               ∠Y = 98.00429185º  

181 градус 02 минуты - 31 градус 08 минуты, как это написать в калькуляторе ?

02 минуты = 0.03333  градус

08 минуты = 0.13333  градус

181.03333 - 31.13333  

Pavlov answered your question, but I'll say it again: the longest side of a right-angled triangle is called the hypotenuse. Squares and rectangles have their hypotenuses, and they are called diagonals.

AB and AC are common tangents from point A to a circle, where B and C lie on the circle.  P is on AB, and R is on AC so that PR is tangent to the circle.  The perimeter of triangle APR is 30, and BC = 10.  Find the perimeter of triangle ABC.

Hint: 

If AP = AR, then PC = 9.357483637

Area of Δ APR = 40.98780306 u²  

Rectangles ABCD and ANME are similar.

AE is 1/2 of AD

AG must be 1/2 of AE