Solutions to the Above Problems

1. If we draw a radius in the small circle to the point of tangency, it will be at right angle with the chord.(see figure below). If x is half the length of AB, r is the radius of the small circle and R the radius of the large circle then by Pythagora's theorem we have:
   
   r² + x² = R²
   
   6² + x² = 10²
   
   Solve for x: x = 8
   
   Length of AB = 2x = 16

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2. Use cosine law in triangle ABE: 13² = 20² + 9² - 2(20)(9)cos(T)
   
   Use cosine law in triangle ACB: x² = 20² + 9² - 2(20)(9)cos(90^o - T)
   

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   Note that cos(90^o - T) = sin(T) and rewrite the second equation as
   
   Use cosine law in triangle ACB: x² = 20² + 9² - 2(20)(9)sin(T)
   
   Solve the first equation for cos(T).
   
   cos(T) = 13/15
   
   Use trigonometric identity to find sin(T) = 2 sqrt(14) / 15
   
   Substitute sin(T) by 2 sqrt(14) / 15 in the third equation and solve for x
   
   x = sqrt( 481 - 48 sqrt( 14 ) ) = 17.4 (approximated to 3 significant digits)

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3. Solve x - y = 1 for x (x = 1 + y) and substitute in the equation of the circle to obtain:
   
   (1 + y)² + 2·(1 + y) + y² + 4y = -1
   
   Write the above quadratic equation in standard form and solve it to obtain
   
   y = - 2 + sqrt(2) and y = - 2 - sqrt(2)
   
   Use x = 1 + y to find x
   
   Points of intersection: ( -1 + sqrt(2), - 2 + sqrt(2) ) and ( -1 - sqrt(2) , -2 - sqrt(2) )

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4. We first graph the lines y = x and y = -2x + 3 in order to locate the points of intersection of the lines and the x axis and identify the triangle in question.
   
   

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   The height is the y coordinate of the point of intersection of the lines y = x and y = -2x + 3 found by solving the system of equations.
   
   solve : y = -2x + 3 , y = x , solution: (1 , 1) which also the point of intersection. The y coordinate = 1 and is also the height.
   
   The length of the base is the x intercept of the line y = -2x + 3 which is x = 3/2.
   
   Area of the shaded triangle = (1/2)(1)(3/2) = 3/4

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5. The formula for the area using two sides and the internal angle they make, may be written as follows
   
   18 = (1/2) * 5 * 10*sin(A)
   
   which gives: sin(A) = 18/25
   
   We now use the cosine formula to fin the length x of the third side opposing angle A as follows:
   
   x² = 5² + 10² - 2*5*10*cos(A)
   
   with cos(A) = sqrt(1 - sin(A)²)
   
   Substitute in the expression for x² and solve for x to obtain x = 7.46 (approximated to 3 significant digits)

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6. The area of triangle BOC is 15 and is given by (1/2) * BO * OC * sin(BOC)
   
   The area of triangle AOD is given by (1/2) * AO * OD * sin(AOD)
   
   Note that angle BOC and AOD are equal.
   
   By the theorem of the intersecting chords we have: AO * OC = BO * OD
   
   Which may be written as: AO / BO = OD / OC = 10 / 5 = 2
   
   The ratios AO / BO and OD / OC are both equal to 2, hence their product is equal to 4 as follows
   
   (AO * OD) / (BO * OC) = 4
   
   Which gives: AO * OD = 4 * (BO * OC)
   
   Hence the area of triangle AOD is 4 times the area of triangle BOC and is equal to 60.
   
   

   

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