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The bearing of P from Q is 072°. Find the bearing of Q from P. A number is selected at random from the set M=2 x < 16. Find the probability that the number is a multiple of 5.

Mathematics

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6. A man 1.6 m tall stands 20 m from a building of height 27 m. Find, correct to one decimal place, the angle of elevation of the top of the building from the man.

  1. Determine the height of the building above the man's eye level.

    Effective height=Building height−Man’s height\text{Effective height} = \text{Building height} - \text{Man's height} Effective height=27 m−1.6 m=25.4 m\text{Effective height} = 27 \text{ m} - 1.6 \text{ m} = 25.4 \text{ m}
  2. Use trigonometry to find the angle of elevation.

    The distance from the building is the adjacent side, and the effective height is the opposite side.

    tan⁡θ=oppositeadjacent\tan \theta = \frac{\text{opposite}}{\text{adjacent}} tan⁡θ=25.420\tan \theta = \frac{25.4}{20} tan⁡θ=1.27\tan \theta = 1.27
  3. Calculate the angle θ\theta.

    θ=arctan⁡(1.27)\theta = \arctan(1.27) θ≈51.78∘\theta \approx 51.78^\circ

    Rounding to one decimal place, θ≈51.8∘\theta \approx 51.8^\circ.

    The correct option is A.

    A. 51.8∘\boxed{\text{A. } 51.8^\circ}

    7. In the diagram, PRPR is tangent to the circle at QQ. ∠SQT=58∘\angle SQT = 58^\circ and ∠STQ=69∘\angle STQ = 69^\circ. Find ∠PQR\angle PQR.

  4. Find ∠QST\angle QST in △SQT\triangle SQT.

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

    ∠QST+∠SQT+∠STQ=180∘\angle QST + \angle SQT + \angle STQ = 180^\circ ∠QST+58∘+69∘=180∘\angle QST + 58^\circ + 69^\circ = 180^\circ ∠QST+127∘=180∘\angle QST + 127^\circ = 180^\circ ∠QST=180∘−127∘=53∘\angle QST = 180^\circ - 127^\circ = 53^\circ
  5. Apply the Alternate Segment Theorem.

    The angle between a tangent (PRPR) and a chord (SQSQ) through the point of contact (QQ) is equal to the angle in the alternate segment (∠QST\angle QST).

    ∠PQR=∠QST\angle PQR = \angle QST ∠PQR=53∘\angle PQR = 53^\circ

    The correct option is A.

    A. 53∘\boxed{\text{A. } 53^\circ}

    8. Given that tan⁡R=724\tan R = \frac{7}{24}, where 0∘<R<90∘0^\circ < R < 90^\circ, find sin⁡R\sin R.

  6. Use the given tangent value to find the hypotenuse of a right-angled triangle.

    tan⁡R=oppositeadjacent=724\tan R = \frac{\text{opposite}}{\text{adjacent}} = \frac{7}{24}

    Let the opposite side be 7 and the adjacent side be 24.
    Using the Pythagorean theorem, the hypotenuse hh is:

    h=opposite2+adjacent2h = \sqrt{\text{opposite}^2 + \text{adjacent}^2} h=72+242h = \sqrt{7^2 + 24^2} h=49+576h = \sqrt{49 + 576} h=625h = \sqrt{625} h=25h = 25
  7. Calculate sin⁡R\sin R.

    sin⁡R=oppositehypotenuse\sin R = \frac{\text{opposite}}{\text{hypotenuse}} sin⁡R=725\sin R = \frac{7}{25}

    The correct option is A.

    A. 725\boxed{\text{A. } \frac{7}{25}}

    9. The bearing of PP from QQ is 072∘072^\circ. Find the bearing of QQ from PP.

  8. Understand the relationship between forward and back bearings.

    The bearing of PP from QQ is 072∘072^\circ. This is the forward bearing.

  9. Calculate the back bearing.

    If the forward bearing is less than 180∘180^\circ, add 180∘180^\circ to find the back bearing.

    Bearing of Q from P=Bearing of P from Q+180∘\text{Bearing of } Q \text{ from } P = \text{Bearing of } P \text{ from } Q + 180^\circ Bearing of Q from P=072∘+180∘\text{Bearing of } Q \text{ from } P = 072^\circ + 180^\circ Bearing of Q from P=252∘\text{Bearing of } Q \text{ from } P = 252^\circ

    The correct option is B.

    B. 252∘\boxed{\text{B. } 252^\circ}

    10. A number is selected at random from the set M={x:2<x<16}M = \{x: 2 < x < 16\}. Find the probability that the number is a multiple of 5.

  10. List the elements in set MM.

    The set MM contains integers strictly greater than 2 and strictly less than 16.

    M={3,4,5,6,7,8,9,10,11,12,13,14,15}M = \{3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}
  11. Count the total number of elements in MM.

    Total number of elements=15−3+1=13\text{Total number of elements} = 15 - 3 + 1 = 13
  12. Identify the multiples of 5 within set MM.

    The multiples of 5 in MM are 5 and 10.

    Number of multiples of 5=2\text{Number of multiples of 5} = 2
  13. Calculate the probability.

    Probability=Number of favorable outcomesTotal number of outcomes\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} Probability=213\text{Probability} = \frac{2}{13}

    The correct option is B.

    B. 213\boxed{\text{B. } \frac{2}{13}}

    11. Given that y=3x2−7x+10y = 3x^2 - 7x + 10, find dydx\frac{dy}{dx} at x=3x=3.

  14. Differentiate yy with respect to xx.

    dydx=ddx(3x2)−ddx(7x)+ddx(10)\frac{dy}{dx} = \frac{d}{dx}(3x^2) - \frac{d}{dx}(7x) + \frac{d}{dx}(10) dydx=3(2x2−1)−7(1x1−1)+0\frac{dy}{dx} = 3(2x^{2-1}) - 7(1x^{1-1}) + 0 dydx=6x−7\frac{dy}{dx} = 6x - 7
  15. Substitute x=3x=3 into the derivative.

    dydx∣x=3=6(3)−7\frac{dy}{dx}\Big|_{x=3} = 6(3) - 7 dydx∣x=3=18−7\frac{dy}{dx}\Big|_{x=3} = 18 - 7 dydx∣x=3=11\frac{dy}{dx}\Big|_{x=3} = 11

    The correct option is B.

    B. 11\boxed{\text{B. } 11}

    12. The interior angle of a regular polygon is 140∘140^\circ. Calculate the sum of the interior angles of the polygon.

  16. Find the number of sides (nn) of the polygon using the formula for an interior angle of a regular polygon.

    Interior angle=(n−2)×180∘n\text{Interior angle} = \frac{(n-2) \times 180^\circ}{n} 140∘=(n−2)×180∘n140^\circ = \frac{(n-2) \times 180^\circ}{n} 140n=180(n−2)140n = 180(n-2) 140n=180n−360140n = 180n - 360 360=180n−140n360 = 180n - 140n 360=40n360 = 40n n=36040n = \frac{360}{40} n=9n = 9

    The polygon has 9 sides.

  17. Calculate the sum of the interior angles using the formula (n−2)×180∘(n-2) \times 180^\circ.

    Sum of interior angles=(9−2)×180∘\text{Sum of interior angles} = (9-2) \times 180^\circ Sum of interior angles=7×180∘\text{Sum of interior angles} = 7 \times 180^\circ Sum of interior angles=1260∘\text{Sum of interior angles} = 1260^\circ

    The correct option is B.

    B. 1260∘\boxed{\text{B. } 1260^\circ}

    13. Given that ab=1p−1q\frac{a}{b} = \frac{1}{p} - \frac{1}{q}, make pp the subject.
    (Assuming the question intended to ask to make pp the subject, as 'C' is not present in the equation and the options are for pp.)

  18. Isolate the term containing pp.

    ab=1p−1q\frac{a}{b} = \frac{1}{p} - \frac{1}{q}

    Add 1q\frac{1}{q} to both sides:

    ab+1q=1p\frac{a}{b} + \frac{1}{q} = \frac{1}{p}
  19. Combine the terms on the left side by finding a common denominator, which is bqbq.

    aqbq+bbq=1p\frac{aq}{bq} + \frac{b}{bq} = \frac{1}{p} aq+bbq=1p\frac{aq + b}{bq} = \frac{1}{p}
  20. Invert both sides to solve for pp.

    p=bqaq+bp = \frac{bq}{aq + b}

    The correct option is B.

    B. p=bqaq+b\boxed{\text{B. } p = \frac{bq}{aq + b}}
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