CBSE Class 10 Mathematics Real Numbers, Algebra, Coordinate Geometry and Trigonometry Assignment Set 16

Answer: (d) 8. The numbers are 2 and 4, so HCF = 2, LCM = 4 and the product is 8.

Teacher's Note:
1. The product of HCF and LCM equals the product of the two numbers.
2. Here 2 × 4 = 8.

Answer: (b) 4. 5005 = 5 × 7 × 11 × 13, which has four prime factors.

Teacher's Note:
1. Divide successively by 5, 7, 11 and 13.
2. All four are prime.

Answer: (a) 2. Zeroes α and 1/α have product 1, and the product is k/2, so k = 2.

Teacher's Note:
1. The product of zeroes of ax² + bx + c is c/a.
2. Set it equal to 1.

Answer: (d) 4. Product of zeroes = −d/a = −(−12)/3 = 4.

Teacher's Note:
1. For a cubic ax³ + bx² + cx + d the product of zeroes is −d/a.
2. Mind the sign.

Answer: (d) no solution. \( \dfrac{3}{12} = \dfrac{4}{16} = \dfrac{1}{4} \neq \dfrac{5}{15} \), so the lines are parallel.

Teacher's Note:
1. Compare the ratios a₁/a₂, b₁/b₂ and c₁/c₂.
2. Two equal ratios and one unequal ratio mean no solution.

Answer: (b) the y-axis. Every point with x = 0 lies on the y-axis.

Teacher's Note:
1. The abscissa is the x-coordinate.
2. All points on the y-axis have x = 0.

Answer: (a) (2, 5). x: (x + (−2))/2 = 0 gives x = 2; y: (y + 3)/2 = 4 gives y = 5.

Teacher's Note:
1. Use the mid-point formula in reverse.
2. Solve for each coordinate separately.

Answer: (a) sin 60°. \( \dfrac{2 \times 1/\sqrt{3}}{1 + 1/3} = \dfrac{2/\sqrt{3}}{4/3} = \dfrac{\sqrt{3}}{2} = \sin 60^\circ \).

Teacher's Note:
1. This is the identity sin 2θ = 2 tan θ/(1 + tan² θ) with θ = 30°.
2. Direct substitution gives √3/2.

Answer: (b) 30°. tan θ = h/(√3 h) = 1/√3, so θ = 30°.

Teacher's Note:
1. Height and shadow form the perpendicular and base of a right triangle.
2. The value of tan 30° is 1/√3.

Answer: (d) tan² A. \( \dfrac{\sec^2 A}{\csc^2 A} = \dfrac{\sin^2 A}{\cos^2 A} = \tan^2 A \).

Teacher's Note:
1. Use 1 + tan² A = sec² A and 1 + cot² A = cosec² A.
2. Then convert to sine and cosine.

Answer: (c) 2. Writing in sin and cos: \( \dfrac{(\cos\theta+\sin\theta+1)(\sin\theta+\cos\theta-1)}{\sin\theta\cos\theta} = \dfrac{(\sin\theta+\cos\theta)^2-1}{\sin\theta\cos\theta} = \dfrac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} = 2 \).

Teacher's Note:
1. Convert every ratio to sine and cosine and use (a + b)(a − b).
2. The product simplifies to the constant 2.

Answer: Let boys = x, so girls = x + 4. Then x + (x + 4) = 10, so 2x = 6 and x = 3. There were 3 boys and 7 girls.

Teacher's Note:
1. Form a linear equation from the total.
2. Check that 3 + 7 = 10 and 7 − 3 = 4.

Answer: Let the x-axis divide AB in the ratio k : 1. The y-coordinate of the point is \( \dfrac{5k-5}{k+1} \), which must be 0, so k = 1. The ratio is 1 : 1 and the point is the mid-point \( \left(\dfrac{1-4}{2}, 0\right) = \left(-\dfrac{3}{2}, 0\right) \).

Teacher's Note:
1. Any point on the x-axis has y = 0.
2. Use this in the section formula.

Answer: Diagonals join (3, 0) to (−1, 4) and (4, 5) to (−2, −1). Lengths: \( d_1 = \sqrt{16+16} = 4\sqrt{2} \) and \( d_2 = \sqrt{36+36} = 6\sqrt{2} \). Area = \( \dfrac{1}{2} d_1 d_2 = \dfrac{1}{2}(4\sqrt{2})(6\sqrt{2}) = 24 \) square units.

Teacher's Note:
1. The area of a rhombus is half the product of its diagonals.
2. Join opposite vertices to get the diagonals.

Answer: Divide the numerator and denominator of the LHS by cos A. Since \( \dfrac{\cot A}{\cos A} = \dfrac{1}{\sin A} = \csc A \), LHS = \( \dfrac{\csc A - 1}{\csc A + 1} \) = RHS.

Teacher's Note:
1. Dividing by cos A converts cot A into cosec A at once.
2. No expansion is needed.

Answer: Let the mid-point be at distance d from each foot. Then \( m_1 = d\tan 60^\circ = \sqrt{3}d \) and \( m_2 = d\tan 30^\circ = \dfrac{d}{\sqrt{3}} \). So \( m_1 : m_2 = \sqrt{3} : \dfrac{1}{\sqrt{3}} = 3 : 1 \).

Teacher's Note:
1. Both statues are at the same horizontal distance from the point.
2. The ratio then equals tan 60° : tan 30°.

Answer: Assume \( 3 + 2\sqrt{5} \) is rational, say \( \dfrac{a}{b} \) with b ≠ 0 and a, b integers. Then \( \sqrt{5} = \dfrac{a - 3b}{2b} \), which is rational. This contradicts the fact that √5 is irrational, so \( 3 + 2\sqrt{5} \) is irrational.

Teacher's Note:
1. Use proof by contradiction and isolate the surd.
2. The sum of a rational number and an irrational multiple is irrational.

Answer: (x − √2)(x + √2) = x² − 2 is a factor. Dividing, \( 2x^4 - 3x^3 - 3x^2 + 6x - 2 = (x^2 - 2)(2x^2 - 3x + 1) \). Now \( 2x^2 - 3x + 1 = (2x - 1)(x - 1) \). The zeroes are \( \sqrt{2}, -\sqrt{2}, 1 \) and \( \dfrac{1}{2} \).

Teacher's Note:
1. Divide the polynomial by the quadratic factor formed from the given zeroes.
2. Factorise the quotient to get the rest.

Answer: Put \( u = \dfrac{1}{x+y} \) and \( v = \dfrac{1}{x-y} \): 10u + 2v = 4 and 15u − 5v = −2. From the first, v = 2 − 5u. Then 15u − 10 + 25u = −2, so u = 1/5 and v = 1. So x + y = 5 and x − y = 1, giving x = 3 and y = 2.

Teacher's Note:
1. Reduce the pair to linear form by substitution of 1/(x + y) and 1/(x − y).
2. Verify in the original equations.

Answer: Let the height be h. Shadow at 30° is \( h\sqrt{3} \) and at 60° is \( \dfrac{h}{\sqrt{3}} \). So \( h\sqrt{3} - \dfrac{h}{\sqrt{3}} = 40 \), i.e. \( \dfrac{2h}{\sqrt{3}} = 40 \) and \( h = 20\sqrt{3} \approx 34.64 \) m.

Teacher's Note:
1. Express each shadow with cot of the altitude.
2. Subtract and equate to 40 m.

Answer: Let each row have x students and there be y rows, so the total is xy. (x + 3)(y − 1) = xy gives x = 3y − 3. (x − 3)(y + 2) = xy gives 2x − 3y = 6. Substituting: 6y − 6 − 3y = 6, so y = 4 and x = 9. Number of students = 9 × 4 = 36.

Teacher's Note:
1. Form two equations from the two conditions.
2. Verify that 12 × 3 = 36 and 6 × 6 = 36.

Answer: The centre is the mid-point (1, 2) and half the diagonal is 2. The other diagonal is perpendicular to the first, so it is vertical through x = 1, and its ends lie 2 units above and below the centre. The other vertices are (1, 4) and (1, 0).

Teacher's Note:
1. Diagonals of a square bisect each other at right angles and are equal.
2. The side check √8 is the same for all sides.

Answer: Since B = 90° − A: tan B = cot A, cot B = tan A, sec B = cosec A and sin B = cos A. Numerator = tan A cot A + tan² A = 1 + tan² A = sec² A. Denominator = sin A cosec A = 1. Also \( \dfrac{\sin^2 B}{\cos^2 A} = 1 \). So LHS = sec² A − 1 = tan² A = RHS.

Teacher's Note:
1. Replace B by 90° − A and use complementary ratios.
2. Then use sec² A − tan² A = 1.

Answer: \( m^2 - n^2 = (m+n)(m-n) = (2\tan\theta)(2\sin\theta) = 4\tan\theta\sin\theta \). Also \( mn = \tan^2\theta - \sin^2\theta = \tan^2\theta(1-\cos^2\theta) = \tan^2\theta\sin^2\theta \), so \( \sqrt{mn} = \tan\theta\sin\theta \). Hence \( m^2 - n^2 = 4\sqrt{mn} \).

Teacher's Note:
1. Factorise m² − n² and simplify mn separately.
2. The two results then match.

Answer: The horizontal distances are \( \dfrac{3600\sqrt{3}}{\tan 60^\circ} = 3600 \) m and \( \dfrac{3600\sqrt{3}}{\tan 30^\circ} = 10800 \) m. Distance covered = 10800 − 3600 = 7200 m in 30 s, i.e. 240 m/s. Speed = 240 × 18/5 = 864 km/h.

Teacher's Note:
1. Find the horizontal position at each instant.
2. Convert m/s to km/h by multiplying by 18/5.