jugoslav

Angle QRC = 15 degrees!!! 

* S i m i l a r  T r i a n g l e s*

AQ => x             QZ => 42 - x

8 / 28 = x / (42 - x)

QZ = 32 ²/₃

The question is incomplete! 

The vertices of triangle ABC lie on the sides of the equilateral triangle DEF, as shown. If CD = 3, CE = BD = 2, and angle ACB = 90º, then find AE.

Oops!    You are right! 

Suppose ABCD is a convex quadrilateral with ABC = 144, ADC = 105, and AB = BD = DC. Compute BCD - BAD.

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Angle ABD = x              

Angle CBD = y

x + y = 144

x = 144 - y

y = 144 - x

∠ADB = 1/2(180 - x)                 (1)

∠BDC = 180 - 2(144 - x)           (2)

(1) + (2) = 105

∠(x) = 82º          ∠(y) = 62º

∠BCD - ∠BAD = 62 - 49

15/8(8 - x) = 8/15(15 - x)

x = 120/23

Holly had a total of 150 books and magazines. She bought another 50 books and gave away 1/3 of her magazines. The number of books and magazines became the same. What was the ratio of the number of books to the number of magazines at first? (Give your answer in the simplest form.)

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b + m = 150          b = 150 - m

b + 50 = m - m/3

150 - m + 50 = m - m/3

m = 120         b = 30

b / m = 1/4

AD = a          AB = b           BD = c

b = sqrt(9 - a²)            

AM = sqrt(a² - 1)            (AM is the height of triangle ADM)

BD * AM = a * b

3[sqrt(a² - 1)] = a[sqrt(9 - a²)]

a = √3          b = √6

[ABCD] = 3√2

Steve and Andrew had $540 altogether. Steve gave 1/7 of his money to Andrew. In return, Andrew gave 1/4 of the total amount he had to Steve. Then they had both an equal amount of money. How much did Steve have at first?

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s + a = 540

s = 540 - a

a + (540 - a)/7 = 360

a = 330

Steve had $210