maths sum, pleeeease help?

Question

Please solve this sum for me, its really challenging for my level, and i just can't get it =(

http://img355.imageshack.us/img355/9581/sumfq2.png

Answer 1

z = 30-60 90 Right triangle

z = 30°

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X = Equalateral Triangle 60 - 60- 60 = 180

x = 60°

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y = right Triangle. Since 40° is given the other angle is 50

y = 50°

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Answer 2

x=50, as x+40 are a right angle (inscribed in a semicircle)
y=50 as x and y are alternate interior angles of parallel lines
z= 40 with a little more work. show the two tall triangles with base on the radii are congruent (SAS), then the vertex angles are both x. The other angle in the left hand small triangle is 90, as it's vertical angle is 90 (after z=40). So, y =50.

Answer 3

i assume the 4 small circles are congruent. enable r be their radius. Then the sq. having their centers as vertices has facets of length 2r. This sq. relies on the midsection of the circle and its diagonal has length sqrt(2)*2r. the area between the midsection of the super circle and the midsection of one of the small circles is a million/2 this diagonal, or sqrt(2)*r The radius of the super circle is (distance from the midsection of the super circle to midsection of small circle + radius of small circle) = sqrt(2)*r + r = 10. fixing for r we get r=10/(sqrt(2)+a million) = 10sqrt(2) - 10. And the radius of a circle equipped in the midsection would be 10-2r = 10 - 20sqrt(2) + 20 = 30 - 20sqrt(2) = a million.7157 cm.

Answer 4

z =40
x = 50

z as all angles are opposite and x because x + z = rightangle

Answer 5

Wow! Me neither!