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Solve the equation: (3X-1)/3 - (2X-1)/4 = 2. Also, find the distance of a landmark from a 30m cliff if the angle of elevation is 60 degrees.

Mathematics

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6. Solve the equation:

3X−13−2X−14=2\frac{3X-1}{3} - \frac{2X-1}{4} = 2

  1. Find a common denominator for the fractions, which is 12.

    Multiply the entire equation by 12.

    12(3X−13)−12(2X−14)=12(2)12 \left( \frac{3X-1}{3} \right) - 12 \left( \frac{2X-1}{4} \right) = 12(2)

  2. Simplify the terms.

    4(3X−1)−3(2X−1)=244(3X-1) - 3(2X-1) = 24

  3. Distribute the numbers into the parentheses.

    12X−4−6X+3=2412X - 4 - 6X + 3 = 24

  4. Combine like terms on the left side.

    (12X−6X)+(−4+3)=24(12X - 6X) + (-4 + 3) = 24 6X−1=246X - 1 = 24

  5. Add 1 to both sides of the equation.

    6X=24+16X = 24 + 1 6X=256X = 25

  6. Divide by 6 to solve for XX.

    X=256X = \frac{25}{6}

    The solution is 256\boxed{\frac{25}{6}}.

    7. The angle of elevation of the top of a cliff 30 metres high from a landmark is 60∘60^\circ. How far is the landmark from the foot of the cliff? Leave your answer in surd form.

  7. Define the variables.

    Let hh be the height of the cliff, h=30 mh = 30 \text{ m}. Let dd be the distance from the landmark to the foot of the cliff. The angle of elevation is θ=60∘\theta = 60^\circ.

  8. Use the tangent function, as it relates the opposite side (height) to the adjacent side (distance) in a right-angled triangle.

    tan⁡(θ)=oppositeadjacent\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} tan⁡(60∘)=hd\tan(60^\circ) = \frac{h}{d}

  9. Substitute the known values.

    We know tan⁡(60∘)=3\tan(60^\circ) = \sqrt{3}.

    3=30d\sqrt{3} = \frac{30}{d}

  10. Solve for dd.

    d=303d = \frac{30}{\sqrt{3}}

  11. Rationalize the denominator by multiplying the numerator and denominator by 3\sqrt{3}.

    d=303×33d = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} d=3033d = \frac{30\sqrt{3}}{3} d=103 md = 10\sqrt{3} \text{ m}

    Comparing this with the options, option (b) is 30/3m30/\sqrt{3} \text{m}, which simplifies to 103m10\sqrt{3} \text{m}.
    The distance is (b) 30/3m\boxed{\text{(b) } 30/\sqrt{3} \text{m}}.

    8. For what value of X is the expression 6X−1X2−X−6\frac{6X-1}{X^2-X-6} not defined?

  12. An algebraic expression is undefined when its denominator is equal to zero.

    Set the denominator to zero.

    X2−X−6=0X^2-X-6 = 0

  13. Factor the quadratic equation.

    We need two numbers that multiply to -6 and add to -1. These numbers are -3 and 2.

    (X−3)(X+2)=0(X-3)(X+2) = 0

  14. Set each factor equal to zero and solve for XX.

    X−3=0  ⟹  X=3X-3 = 0 \implies X = 3 X+2=0  ⟹  X=−2X+2 = 0 \implies X = -2

    The expression is not defined for X=3X=3 and X=−2X=-2.
    The correct option is (b) 3,−2\boxed{\text{(b) } 3, -2}.

    9. AOC is the diameter of a circle, if ∠ADB=32∘\angle ADB = 32^\circ, calculate ∠CAB\angle CAB.

  15. Identify properties of angles in a circle.

    Since AOC is the diameter, the angle subtended by the diameter at any point on the circumference is a right angle. Therefore, ∠ABC=90∘\angle ABC = 90^\circ.

  16. Angles subtended by the same arc at the circumference are equal.

    ∠ADB\angle ADB and ∠ACB\angle ACB are both subtended by arc AB.

    ∠ACB=∠ADB=32∘\angle ACB = \angle ADB = 32^\circ

  17. Consider △ABC\triangle ABC.

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

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

  18. Substitute the known angles into the equation.

    ∠CAB+90∘+32∘=180∘\angle CAB + 90^\circ + 32^\circ = 180^\circ ∠CAB+122∘=180∘\angle CAB + 122^\circ = 180^\circ

  19. Solve for ∠CAB\angle CAB.

    ∠CAB=180∘−122∘\angle CAB = 180^\circ - 122^\circ ∠CAB=58∘\angle CAB = 58^\circ

    The correct option is (d) 58∘\boxed{\text{(d) } 58^\circ}.

    10. ABCD is a cyclic quadrilateral. If ∠ABD=18∘\angle ABD = 18^\circ, ∠ACB=40∘\angle ACB = 40^\circ, calculate ∠DAB\angle DAB.

  20. Identify properties of angles in a circle.

    Angles subtended by the same arc at the circumference are equal.

    • ∠ADB\angle ADB and ∠ACB\angle ACB are both subtended by arc AB. Therefore, ∠ADB=∠ACB=40∘\angle ADB = \angle ACB = 40^\circ.
    • ∠ACD\angle ACD and ∠ABD\angle ABD are both subtended by arc AD. Therefore, ∠ACD=∠ABD=18∘\angle ACD = \angle ABD = 18^\circ.

  21. Consider △ABD\triangle ABD.

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

    ∠DAB+∠ABD+∠ADB=180∘\angle DAB + \angle ABD + \angle ADB = 180^\circ

  22. Substitute the known angles into the equation.

    ∠DAB+18∘+40∘=180∘\angle DAB + 18^\circ + 40^\circ = 180^\circ ∠DAB+58∘=180∘\angle DAB + 58^\circ = 180^\circ

  23. Solve for ∠DAB\angle DAB.

    ∠DAB=180∘−58∘\angle DAB = 180^\circ - 58^\circ ∠DAB=122∘\angle DAB = 122^\circ

    The correct option is (d) 122∘\boxed{\text{(d) } 122^\circ}.

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