CBSE Class 10 Mathematics Trigonometry, Polynomials, Quadratic Equations and Statistics Assignment Set 10

Answer: sin 30° = 1/2 = cos 60°, so 40° + x = 60° and x = 20°.

Teacher's Note:
1. Express sin 30° as a cosine of the same type.
2. Equate the angles since both are acute.

Answer: t² − 15 = 0 gives t² = 15, so the zeroes are √15 and −√15.

Teacher's Note:
1. Use the identity a² − b² = (a − b)(a + b) if you like.
2. Write both signs of the square root.

Answer: Arranged in order: 3, 5, 7, 8, 9, 10, 14, 17, 23. There are 9 observations, so the median is the 5th value, 9.

Teacher's Note:
1. Always arrange the data in ascending order first.
2. For odd n the median is the ((n + 1)/2)th term.

Answer: cos 90° = 0, so 4x − 10° = 90°, giving 4x = 100° and x = 25°.

Teacher's Note:
1. Recall that cos 90° = 0.
2. Solve the linear equation in x.

Answer: Substituting x = 8: 64 − 80 + k = 0, so k = 16.

Teacher's Note:
1. A root satisfies the equation.
2. Check: x² − 10x + 16 has roots 2 and 8.

Answer: Adding: 100(x + y) = 1000, so x + y = 10. Subtracting: 2x − 2y = 2, so x − y = 1.
Hence x = 5.5 and y = 4.5.

Teacher's Note:
1. Equations with swapped coefficients are solved by adding and subtracting.
2. Verify in the first equation: 269.5 + 229.5 = 499.

Answer: tan θ = 4/5, so the hypotenuse is √(16 + 25) = √41, sin θ = 4/√41 and cos θ = 5/√41.
5 sin θ − 3 cos θ = (20 − 15)/√41 = \( \dfrac{5}{\sqrt{41}} \).

Teacher's Note:
1. Draw a right triangle with perpendicular 4 and base 5.
2. Find the hypotenuse using the Pythagoras theorem.

Answer: α + β = 2 and αβ = 5/2. So α² + β² = 4 − 5 = −1.
Value = (α² + β²)/(αβ)² = −1/(25/4) = \( -\dfrac{4}{25} \).

Teacher's Note:
1. Write the expression as (α² + β²)/(αβ)².
2. Use the sum and product of zeroes formulae.

Answer: The number divides 70 − 5 = 65 and 125 − 8 = 117. 65 = 5 × 13 and 117 = 9 × 13, so HCF = 13.
The largest number is 13.

Teacher's Note:
1. Subtract the remainders first and then find the HCF.
2. The divisor must also be larger than the remainders, and 13 is.

Answer: Let 1 man do 1/x of the work per day and 1 boy 1/y. Then 4/x + 6/y = 1/5 and 3/x + 4/y = 1/7. Put u = 1/x, v = 1/y: 4u + 6v = 1/5 and 3u + 4v = 1/7.
Solving gives u = 1/35 and v = 1/70. So 1 man takes 35 days and 1 boy takes 70 days.

Teacher's Note:
1. Use daily work rates as the variables.
2. Check: 4/35 + 6/70 = 1/5.

Answer: Let the speeds be x km/h (train) and y km/h (car). 320/x + 280/y = 26/3 and 200/x + 400/y = 26/3 + 1/2 = 55/6.
With u = 1/x, v = 1/y, solving gives u = 1/80, v = 1/60. So the train's speed is 80 km/h and the car's speed is 60 km/h.

Teacher's Note:
1. Convert 8 hours 40 minutes to 26/3 hours.
2. Verify: 320/80 + 280/60 = 4 + 14/3 = 26/3.

Answer: LHS = sin A + sin²A/cos A + cos A + cos²A/sin A = (sin A + cos A) + (sin³A + cos³A)/(sin A cos A).
Using sin³A + cos³A = (sin A + cos A)(1 − sin A cos A), LHS = (sin A + cos A)[1 + (1 − sin A cos A)/(sin A cos A)] = (sin A + cos A)/(sin A cos A) = cosec A + sec A = RHS.

Teacher's Note:
1. Convert tan and cot to sine and cosine.
2. Use the identity for the sum of cubes.

Answer: LHS = 2 tan²θ + tan θ + 4 tan θ + 2 = 2 tan²θ + 5 tan θ + 2 = 5 tan θ + 2(tan²θ + 1) = 5 tan θ + 2 sec²θ = RHS.

Teacher's Note:
1. Expand the product first.
2. Then use 1 + tan²θ = sec²θ.

Answer: m² = 1 + 2 sin θ cos θ, so m² − 1 = 2 sin θ cos θ. Also n = (sin θ + cos θ)/(sin θ cos θ) = m/(sin θ cos θ).
Hence n(m² − 1) = [m/(sin θ cos θ)] × 2 sin θ cos θ = 2m.

Teacher's Note:
1. Express m² − 1 and n in terms of sin θ cos θ.
2. Then multiply them.

Answer: Let the zeroes be α and 2α. Then 3α = −b/a and 2α² = c/a. From the first, α = −b/(3a), so 2b²/(9a²) = c/a, that is 2b² = 9ac.

Teacher's Note:
1. Use sum and product of zeroes.
2. Eliminate α between the two relations.

Answer: Sum = 0, sum of products in pairs = −12 − 4 + 3 = −13, product = 12.
The polynomial is x³ − 13x − 12 (or any non-zero multiple).

Teacher's Note:
1. Use x³ − (sum)x² + (pair sum)x − product.
2. Check by putting x = 4: 64 − 52 − 12 = 0.

Answer: p(−1) = k + 4 + k = 0, so 2k = −4 and k = −2.

Teacher's Note:
1. A zero makes the polynomial value 0.
2. Substitute x = −1 carefully with signs.

Answer: Let the angles be θ and 90° − θ and the height h. Then tan θ = h/a and tan (90° − θ) = cot θ = h/b.
Multiplying, tan θ cot θ = 1 = h²/(ab), so h² = ab and h = √(ab).

Teacher's Note:
1. Use tan θ · cot θ = 1 for complementary angles.
2. The two points are on the same side of the tower.

Answer: x + y = 60 − 45 = 15. The median class is 20-30 with l = 20, f = 20, cf = 5 + x, h = 10.
20 + [(30 − 5 − x)/20] × 10 = 28.5, so 25 − x = 17 and x = 8, y = 7.

Teacher's Note:
1. Find the median class using n/2 = 30.
2. Verify that the cumulative frequency 33 reaches the median class.

Answer: Let each side be a. Draw AE ⊥ BC; E is the midpoint, so BE = a/2, AE² = 3a²/4 and BD = a/3, so DE = a/2 − a/3 = a/6.
AD² = AE² + DE² = 3a²/4 + a²/36 = 28a²/36 = 7a²/9, so 9AD² = 7a² = 7AB².

Teacher's Note:
1. Use the altitude of an equilateral triangle bisecting the base.
2. Apply Pythagoras in the right triangle ADE.

Answer: Together they fill in 35/12 h. Let the larger tap take x hours and the smaller x + 2. Then 1/x + 1/(x + 2) = 12/35, so 35(2x + 2) = 12x(x + 2), which gives 6x² − 23x − 35 = 0.
x = (23 + 37)/12 = 5 (the negative root is rejected). The larger tap takes 5 hours and the smaller 7 hours.

Teacher's Note:
1. Add the rates of the two taps to get the combined rate.
2. Reject the negative value of time.

Answer: The standing part is 10 tan 45° = 10 m and the broken part is 10/cos 45° = 10√2 m.
Height of the tree = 10 + 10√2 = 10(1 + √2) ≈ 24.14 m.

Teacher's Note:
1. The original height is the standing part plus the bent part.
2. The bent part is the hypotenuse of the right triangle.

Answer: The sum of the coefficients is (b − c) + (c − a) + (a − b) = 0, so x = 1 is a root. Since the roots are equal, both roots are 1, so the product of roots is (a − b)/(b − c) = 1.
So a − b = b − c, that is 2b = a + c.

Teacher's Note:
1. Notice that x = 1 satisfies the equation.
2. Equal roots means the other root is also 1.

Answer: Horizontal distance at 60°: 3000√3/tan 60° = 3000 m. At 30°: 3000√3/tan 30° = 9000 m.
Distance covered = 6000 m in 30 s, so speed = 200 m/s = 720 km/h.

Teacher's Note:
1. Find the horizontal distance for each angle.
2. Speed equals the difference of distances divided by time.

Answer: m² − n² = (m + n)(m − n) = (2 tan θ)(2 sin θ) = 4 tan θ sin θ.
mn = tan²θ − sin²θ = tan²θ(1 − cos²θ) = tan²θ sin²θ, so √(mn) = tan θ sin θ. Hence m² − n² = 4√(mn).

Teacher's Note:
1. Factorise m² − n² and find mn separately.
2. Use tan²θ − sin²θ = tan²θ(1 − cos²θ).