Dragan

Great job, geno!!!

in triangle ABC, AB = AC and D is a point on line AC so that line BD bisects angle ABC. if BD = BC what is the measure, in degrees, of angle A?

Let 1/2 of ∠ABC be an x

Hint:   The sum of angles in a triangle BCD is  5x, and ∠A = x  

That can be done this way: multiply decimal number by a whole number in order to get a whole number as a product.

( Hope it's understandable. English is my second language.)

9.428571429 * 5 =  no

9.428571429 * 6 =  no

9.428571429 * 7 =  66  

Maybe there's some more "mathematical" way to do that.

Hello, ilorty! Thanks for defending me so zealously!!!

I'm that Guest above your post; number in red is wrong! I copied it off Noori's post. 

(btw,  Dragan is a common male name in my homeland. People often confuse my name with dragon)

***Arc  SR is not 106º ***      it's  117.52º   ( you don't need it, anyway)

Since the points S, T, and R are tangent points, we have:

         ET = SE = 26        |         SD = DR = 32        |       RC = TC = 48

                 ( 26                +                 32                 +                48 ) * 2  = Perimeter  

Hint:   triangle PSU is an equilateral triangle.

Use the law of cosines to calculate the angels.

a² = b² + c² −2bc * cos(A)

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Example:

Triangle ABC has sides: a=2, b=3, c=4

2² = 3² + 4² - 2*3*4*cos(A)

cos(A) = 3² + 4² - 2²/ 2*3*4 = 0.875

Angle  A = 28.955º

This is how that inscribed triangle actually looks.

Scale 1:1  

1)   In the diagram, angle U = 30 degrees. Arc XY is 170 degrees, and arc VW is 110 degrees. Find arc WY, in degrees.

Hint:   line segment VY is a diameter of a circle; the sum of arc VX and arc WY is 80 degrees.

∠U = 1/2 ( arcWY - arcVX )

 Arc WY - Arc XV = Angle U = 30 degrees.

Arc WY + Arx XV = 360 - 170 - 110 = 80 degrees.

Therefore, arc WY = (30 + 80)/2 = 55 degrees.

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This is the answer to question #1 from that link; the answer is not correct!!!