CBSE Class 10 Mathematics Real Numbers, Polynomials, Linear Equations, Triangles and Trigonometry Worksheet Set 08

Answer: The required number divides 70 − 5 = 65 and 125 − 8 = 117 exactly.
65 = 5 × 13 and 117 = 3² × 13, so HCF(65, 117) = 13.
13 is greater than both remainders (5 and 8), so it is valid. Answer: 13.

Teacher's Note:
1. Subtract the remainders first so that the numbers become exactly divisible.
2. The divisor must also be larger than each remainder, which 13 satisfies.

Answer: Suppose \( \sqrt{5} = \dfrac{p}{q} \), where p and q are co-prime integers and q ≠ 0. Then 5q² = p², so 5 divides p², hence 5 divides p. Let p = 5m. Then 5q² = 25m², so q² = 5m², and 5 divides q as well. Now 5 is a common factor of p and q, which contradicts the assumption that they are co-prime. So \( \sqrt{5} \) is irrational.
Next, suppose \( 3+\sqrt{5} = r \) is rational. Then \( \sqrt{5} = r-3 \), which is rational, because the difference of two rational numbers is rational. This contradicts the first result, so \( 3+\sqrt{5} \) is irrational.

Teacher's Note:
1. Proof by contradiction is the standard method for irrational numbers.
2. In the second part, a rational number plus or minus a rational number is always rational.

Answer: \( t^2-15 = (t-\sqrt{15})(t+\sqrt{15}) \). Putting this equal to 0 gives the zeroes \( \sqrt{15} \) and \( -\sqrt{15} \).

Teacher's Note:
1. Write the polynomial as a difference of squares and factorise.
2. Each factor set to zero gives one zero.

Answer: α + β = −(−4)/2 = 2 and αβ = 5/2.
\( \dfrac{1}{\alpha^2}+\dfrac{1}{\beta^2} = \dfrac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2} = \dfrac{4-5}{25/4} = \dfrac{-1\times 4}{25} \).
Answer: \( -\dfrac{4}{25} \).

Teacher's Note:
1. Express the required sum through α + β and αβ before substituting.
2. The zeroes here are not real, but the algebraic relations still hold.

Answer: Let the zeroes be α and 2α. Then sum: 3α = −b/a, so α = −b/(3a). Product: 2α² = c/a.
Substituting α: 2 × b²/(9a²) = c/a, so 2b² = 9ac. Hence proved.

Teacher's Note:
1. Use both the sum and the product of zeroes and eliminate α.
2. Cross-multiplying 2b²/(9a²) = c/a gives the result.

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

Teacher's Note:
1. If a number is a zero, substituting it makes the polynomial equal to 0.
2. Take care with the sign of the −4x term when x = −1.

Answer: Adding the equations: 100x + 100y = 1000, so x + y = 10.
Subtracting the first from the second: 2x − 2y = 2, so x − y = 1.
Solving: x = 5.5 and y = 4.5. Check: 49(5.5) + 51(4.5) = 269.5 + 229.5 = 499.

Teacher's Note:
1. When coefficients are interchanged, adding and subtracting the equations simplifies the work.
2. Always substitute back to check both equations.

Answer: Let the numbers be x and y. Then x + y = 18 and x − y = 6. Adding: 2x = 24, so x = 12 and y = 6. The numbers are 12 and 6.

Teacher's Note:
1. Form two equations from the sum and the difference.
2. Adding them removes y immediately.

Answer: Since 8 satisfies the equation, 64 − 80 + k = 0, so k = 16.

Teacher's Note:
1. A root of an equation satisfies it, so substitute it directly.
2. The other root here is 2, since 8 + 2 = 10.

Answer: Since the triangle is isosceles and right-angled at C, AC = BC.
By the Pythagoras theorem, AB² = AC² + BC² = BC² + BC² = 2BC². Hence proved.

Teacher's Note:
1. The equal sides of a right isosceles triangle are the two sides forming the right angle.
2. Replace AC with BC in the Pythagoras relation.

Answer: Let each side be a, so BD = a/3. Draw AE ⟂ BC. E is the midpoint of BC, so BE = a/2 and DE = a/2 − a/3 = a/6.
In right triangle AEB: AE² = a² − a²/4 = 3a²/4.
In right triangle AED: AD² = AE² + DE² = 3a²/4 + a²/36 = 28a²/36 = 7a²/9.
So 9AD² = 7a² = 7AB². Hence proved.

Teacher's Note:
1. The altitude of an equilateral triangle meets the base at its midpoint.
2. Apply Pythagoras in two right triangles, AEB and AED.

Answer: For similar triangles ABC and PQR, area(ABC)/area(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)². The areas are equal, so this ratio is 1.
Hence AB = PQ, BC = QR and CA = RP. By the SSS criterion, the triangles are congruent.

Teacher's Note:
1. The ratio of areas of similar triangles equals the square of the ratio of corresponding sides.
2. A ratio of 1 means all corresponding sides are equal.

Answer: sin 30° = 1/2 = cos 60°. So cos (40° + x) = cos 60°, which gives 40° + x = 60° and x = 20°.

Teacher's Note:
1. Convert sin 30° to a cosine value of a standard angle.
2. Equal cosines of acute angles imply equal angles.

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

Teacher's Note:
1. The only acute angle with cosine 0 in the standard table is 90°.
2. Solve the resulting linear equation.

Answer: tan θ = 4/5, so take opposite = 4k and adjacent = 5k. Hypotenuse = \( \sqrt{16+25}\,k=\sqrt{41}\,k \).
sin θ = \( \dfrac{4}{\sqrt{41}} \) and cos θ = \( \dfrac{5}{\sqrt{41}} \).
5 sin θ − 3 cos θ = \( \dfrac{20-15}{\sqrt{41}} = \dfrac{5}{\sqrt{41}} \).

Teacher's Note:
1. Draw a right triangle from the given tangent and find the hypotenuse.
2. Substitute sin θ and cos θ and simplify.

Answer: LHS = sin A (cos A + sin A)/cos A + cos A (sin A + cos A)/sin A
= (sin A + cos A) [sin A/cos A + cos A/sin A]
= (sin A + cos A)(sin² A + cos² A)/(sin A cos A)
= (sin A + cos A)/(sin A cos A) = 1/cos A + 1/sin A = sec A + cosec A = RHS.

Teacher's Note:
1. Convert tan and cot into sine and cosine and take (sin A + cos A) common.
2. The identity sin² A + cos² A = 1 does the rest.

Answer: LHS = 2tan² θ + tan θ + 4 tan θ + 2 = 2tan² θ + 5 tan θ + 2
= 2(tan² θ + 1) + 5 tan θ = 2 sec² θ + 5 tan θ = RHS.

Teacher's Note:
1. Expand the product first and then group 2tan² θ + 2.
2. Use 1 + tan² θ = sec² θ.

Answer: m² = sin² θ + cos² θ + 2 sin θ cos θ = 1 + 2 sin θ cos θ, so m² − 1 = 2 sin θ cos θ.
n = 1/cos θ + 1/sin θ = (sin θ + cos θ)/(sin θ cos θ) = m/(sin θ cos θ).
n(m² − 1) = [m/(sin θ cos θ)] × 2 sin θ cos θ = 2m. Hence proved.

Teacher's Note:
1. Square m to obtain an expression for m² − 1.
2. Write n as a single fraction and multiply.

Answer: m² − n² = (m + n)(m − n) = (2 tan θ)(2 sin θ) = 4 tan θ sin θ.
mn = tan² θ − sin² θ = sin² θ (1/cos² θ − 1) = sin² θ tan² θ, so √(mn) = sin θ tan θ.
Hence m² − n² = 4√(mn).

Teacher's Note:
1. First find m + n and m − n, then find mn separately.
2. Factor sin² θ out of tan² θ − sin² θ.

Answer: Let the height be h. From the nearer point at distance a, tan θ = h/a. From the other point at distance b, the angle is 90° − θ, so 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. Complementary angles have tangents that are reciprocals of each other.
2. Multiplying the two tangent relations eliminates the angle.

Answer: Standing part = 10 tan 45° = 10 m. Broken part (hypotenuse) = 10/cos 45° = 10√2 m.
Original height = 10 + 10√2 = 10(1 + √2) m ≈ 24.14 m.

Teacher's Note:
1. The tree is split into a vertical part and a slanting part, and both must be added.
2. Use tan 45° for the vertical part and cos 45° for the slanting part.

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

Teacher's Note:
1. Find the horizontal distance of the plane at both instants and subtract.
2. Divide the distance flown by the time to get speed.