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Find the value of y in the figure below, given that O is the centre of the circle.

Mathematics

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3. Find the value of yy in the figure below, given that O is the centre of the circle.

  1. Identify the relationship between the angles.

    In a circle, the angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the remaining part of the circle (circumference).
    From the diagram, the angle at the center is 55∘55^\circ (labeled at O), and the angle at the circumference is y∘y^\circ (labeled at P). However, based on the options provided, it is highly probable that the 55∘55^\circ is the angle at the circumference (∠QPR\angle QPR) and yy is the angle at the center (∠QOR\angle QOR). We will proceed with this common interpretation for such problems when options are integers.

  2. Apply the angle relationship.

    If the angle at the circumference ∠QPR=55∘\angle QPR = 55^\circ, then the angle at the center ∠QOR\angle QOR is twice this value.

    y=2×55∘y = 2 \times 55^\circ y=110∘y = 110^\circ

    This corresponds to option B.

    The final answer is B. 110∘\boxed{\text{B. } 110^\circ}.

    4. In the figure below, AD is a tangent to the circle ABC, ∣AB∣=∣AC∣|AB| = |AC| and ∠BAC=2x∘\angle BAC = 2x^\circ. Find ∠DAC\angle DAC.

  3. Use the property of an isosceles triangle.

    Given that ∣AB∣=∣AC∣|AB| = |AC|, triangle ABC is an isosceles triangle. Therefore, the base angles are equal: ∠ABC=∠ACB\angle ABC = \angle ACB.

  4. Find the measure of ∠ABC\angle ABC.

    The sum of angles in a triangle is 180∘180^\circ.

    ∠BAC+∠ABC+∠ACB=180∘\angle BAC + \angle ABC + \angle ACB = 180^\circ

    Substitute ∠BAC=2x∘\angle BAC = 2x^\circ and ∠ACB=∠ABC\angle ACB = \angle ABC:

    2x∘+∠ABC+∠ABC=180∘2x^\circ + \angle ABC + \angle ABC = 180^\circ 2x∘+2∠ABC=180∘2x^\circ + 2\angle ABC = 180^\circ 2∠ABC=180∘−2x∘2\angle ABC = 180^\circ - 2x^\circ ∠ABC=180∘−2x∘2\angle ABC = \frac{180^\circ - 2x^\circ}{2} ∠ABC=(90−x)∘\angle ABC = (90 - x)^\circ
  5. Apply the Alternate Segment Theorem.

    The Alternate Segment Theorem states that the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.
    Here, AD is the tangent and AC is the chord. The angle in the alternate segment is ∠ABC\angle ABC.
    Therefore, ∠DAC=∠ABC\angle DAC = \angle ABC.

    ∠DAC=(90−x)∘\angle DAC = (90 - x)^\circ

    This corresponds to option B.

    The final answer is B. (90−x)∘\boxed{\text{B. } (90 - x)^\circ}.

    5. A fair die is rolled once. What is the probability of obtaining a 3 or 4?

  6. Determine the total number of possible outcomes.

    A fair die has 6 faces, so the possible outcomes are {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}.
    Total number of outcomes = 6.

  7. Determine the number of favorable outcomes.

    We want to obtain a 3 or a 4. The favorable outcomes are {3,4}\{3, 4\}.
    Number of favorable outcomes = 2.

  8. Calculate the probability.

    The probability of an event is given by the formula:

    P(event)=Number of favorable outcomesTotal number of possible outcomesP(\text{event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} P(3 or 4)=26P(\text{3 or 4}) = \frac{2}{6}
  9. Simplify the probability.

    P(3 or 4)=13P(\text{3 or 4}) = \frac{1}{3}

    This corresponds to option D.

    The final answer is D. 13\boxed{\text{D. } \frac{1}{3}}.

    6. Find the midpoint of a line joining the points A(2, 2) and B(6, 4).

  10. Recall the midpoint formula.

    For two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), the midpoint (xm,ym)(x_m, y_m) is given by:

    (xm,ym)=(x1+x22,y1+y22)(x_m, y_m) = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)
  11. Substitute the coordinates of points A and B.

    Given A(2,2)A(2, 2) and B(6,4)B(6, 4), we have x1=2,y1=2,x2=6,y2=4x_1=2, y_1=2, x_2=6, y_2=4.

    xm=2+62x_m = \frac{2+6}{2} ym=2+42y_m = \frac{2+4}{2}
  12. Calculate the coordinates of the midpoint.

    xm=82=4x_m = \frac{8}{2} = 4 ym=62=3y_m = \frac{6}{2} = 3

    The midpoint is (4,3)(4, 3).
    This corresponds to option A.

    The final answer is A. (4,3)\boxed{\text{A. } (4, 3)}.

    7. Find the distance between the points A(3, -3) and B(4, -3).

  13. Recall the distance formula.

    For two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), the distance dd is given by:

    d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}
  14. Substitute the coordinates of points A and B.

    Given A(3,−3)A(3, -3) and B(4,−3)B(4, -3), we have x1=3,y1=−3,x2=4,y2=−3x_1=3, y_1=-3, x_2=4, y_2=-3.

    d=(4−3)2+(−3−(−3))2d = \sqrt{(4-3)^2 + (-3 - (-3))^2}
  15. Simplify the expression.

    d=(1)2+(−3+3)2d = \sqrt{(1)^2 + (-3+3)^2} d=12+02d = \sqrt{1^2 + 0^2} d=1+0d = \sqrt{1 + 0} d=1d = \sqrt{1} d=1d = 1

    This corresponds to option D.

    The final answer is D. 1\boxed{\text{D. } 1}.

    8. If the equation 4x2−4Px+3=04x^2 - 4\sqrt{P}x + 3 = 0 has equal roots, find the value of PP.

  16. Recall the condition for equal roots in a quadratic equation.

    A quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0 has equal roots if its discriminant (Δ\Delta) is equal to zero.

    Δ=b2−4ac=0\Delta = b^2 - 4ac = 0
  17. Identify the coefficients of the given equation.

    The equation is 4x2−4Px+3=04x^2 - 4\sqrt{P}x + 3 = 0.
    Comparing this to ax2+bx+c=0ax^2 + bx + c = 0, we have:
    a=4a = 4
    b=−4Pb = -4\sqrt{P}
    c=3c = 3

  18. Substitute the coefficients into the discriminant formula and solve for PP.

    (−4P)2−4(4)(3)=0(-4\sqrt{P})^2 - 4(4)(3) = 0 (16P)−48=0(16P) - 48 = 0 16P=4816P = 48 P=4816P = \frac{48}{16} P=3P = 3

    This corresponds to option C.

    The final answer is C. 3\boxed{\text{C. } 3}.

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