CBSE Class 10 Mathematics Trigonometry, Circles, Areas, Surface Areas and Volumes, Statistics and Probability Assignment Set 12

Answer: cos(90° − 3A) = cos(A − 26°), so 90° − 3A = A − 26°.
4A = 116°, so A = 29°.

Teacher's Note:
1. Convert sin 3A into a cosine using the complementary angle identity.
2. Equal cosines of acute angles mean equal angles.

Answer: Since 65° = 90° − 25°, cos 65° = sin 25° and sin 65° = cos 25°.
The expression is sin²25° + cos²25° = 1.

Teacher's Note:
1. Use the complementary angle relations.
2. Then apply sin²θ + cos²θ = 1.

Answer: x² = a²sin²A + 2ab sin A cos A + b²cos²A
y² = a²cos²A − 2ab sin A cos A + b²sin²A
Adding, the middle terms cancel: x² + y² = a²(sin²A + cos²A) + b²(cos²A + sin²A) = a² + b². Hence proved.

Teacher's Note:
1. Square both expressions and add.
2. The cross terms have opposite signs and cancel.

Answer: The value of cos A decreases from 1 to 0.
For example, cos 0° = 1, cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2 and cos 90° = 0.

Teacher's Note:
1. Compare the standard values of cos at 0°, 30°, 45°, 60° and 90°.
2. Sine behaves in the opposite way.

Answer: tan θ = height/shadow = 1.
So θ = 45°.

Teacher's Note:
1. The height and the shadow are the perpendicular and the base of a right triangle.
2. tan 45° = 1.

Answer: Height of the tower above the eye level = 22 − 1.5 = 20.5 m.
tan θ = 20.5/20.5 = 1, so θ = 45°.

Teacher's Note:
1. Subtract the observer's height from the tower's height.
2. The horizontal distance equals the remaining height.

Answer: Let PA and PB be tangents from P to a circle with centre O, touching it at A and B. Join OA, OB and OP.
OA ⊥ PA and OB ⊥ PB, since a tangent is perpendicular to the radius at the point of contact.
In right triangles OAP and OBP: OA = OB (radii), OP is common, and the angles at A and B are 90°. So the triangles are congruent (RHS) and PA = PB. Hence proved.

Teacher's Note:
1. Use the fact that the radius is perpendicular to the tangent.
2. RHS congruence of two right triangles gives the result.

Answer: The radius OP is perpendicular to the tangent PQ, so triangle OPQ is right-angled at P.
OP² = OQ² − PQ² = 169 − 144 = 25, so OP = 5 cm.

Teacher's Note:
1. Join the centre to the point of contact.
2. Apply Pythagoras theorem in the right triangle.

Answer: Let the circle touch AB at F and AC at E, and let AF = AE = x (tangents from A). Then AB = x + 8, AC = x + 6, BC = 14, and the semi-perimeter s = x + 14.
Area = r × s = 4(x + 14). By Heron's formula the area is √[(x + 14)(x)(8)(6)] = √[48x(x + 14)].
Squaring: 16(x + 14)² = 48x(x + 14), so x + 14 = 3x and x = 7.
AB = 15 cm and AC = 13 cm. Check: s = 21, area = 84 = 4 × 21.

Teacher's Note:
1. Tangents from one point are equal, so the sides can be written using one unknown.
2. Area of a triangle with an incircle = r × semi-perimeter.

Answer: πr² = 2πr gives r = 2.
The diameter is 2r = 4 units.

Teacher's Note:
1. Equate the two formulas and cancel πr.
2. The diameter is twice the radius.

Answer: Area of the sector = \( \dfrac{60}{360} \times \dfrac{22}{7} \times 14^2 \) = 102.67 cm².
Area of the equilateral triangle formed (side 14) = \( \dfrac{\sqrt{3}}{4} \times 14^2 \) = 49√3 = 84.77 cm².
Minor segment = 102.67 − 84.77 ≈ 17.9 cm².
Area of the circle = \( \dfrac{22}{7} \times 196 \) = 616 cm². Major segment = 616 − 17.9 ≈ 598.1 cm².

Teacher's Note:
1. A 60° sector with equal radii forms an equilateral triangle.
2. Major segment = area of circle − minor segment.

Answer: The major sector has angle 360° − 120° = 240°.
Difference = \( \dfrac{240-120}{360} \times \dfrac{22}{7} \times 21^2 \) = \( \dfrac{1}{3} \times 22 \times 3 \times 21 \) = 462 cm².

Teacher's Note:
1. Subtract the angles first, then find the sector area once.
2. 21² × 22/7 = 1386, and one third of that is 462.

Answer: r = 7 cm. Volume = \( \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \dfrac{22}{7} \times 49 \times 51 \) = 2618 cm³.
Mass = volume × density = 2618 × 10 = 26180 g = 26.18 kg.

Teacher's Note:
1. Use the radius, not the diameter, in the volume formula.
2. Mass = volume × density.

Answer: The cuboid is 20 cm × 20 cm × 80 cm.
Surface area = 2(lb + bh + hl) = 2(400 + 1600 + 1600) = 2 × 3600 = 7200 cm².

Teacher's Note:
1. Stacking gives length and breadth 20 cm and height 80 cm.
2. The joined faces are inside and are not counted.

Answer: Mode = 3 × median − 2 × mean = 3(9.6) − 2(10.5) = 28.8 − 21 = 7.8.

Teacher's Note:
1. Use the empirical relation between mean, median and mode.
2. The mode is smaller than the median here because the mean is greater.

Answer: More than 100: 50; more than 120: 38; more than 140: 24; more than 160: 16; more than 180: 10.
The points to plot are (100, 50), (120, 38), (140, 24), (160, 16), (180, 10), together with (200, 0). Join them by a smooth curve.

Teacher's Note:
1. For a more-than table, start from the total and subtract each class frequency in turn.
2. Plot cumulative frequency against the lower class limit.

Answer: Multiples of 2: 10 cards; multiples of 3: 6 cards; multiples of 6 (counted twice): 3 cards.
Favourable outcomes = 10 + 6 − 3 = 13.
P = 13/20.

Teacher's Note:
1. Subtract the numbers that are counted twice.
2. Total outcomes are 20.

Answer: Total outcomes = 4 × 4 = 16.
Product less than 16: x = 1 with y = 1, 4, 9 (3 ways); x = 2 with y = 1, 4 (2 ways); x = 3 with y = 1, 4 (2 ways); x = 4 with y = 1 (1 way). Favourable = 8.
P = 8/16 = 1/2.

Teacher's Note:
1. List the pairs systematically for each value of x.
2. 16 is not less than 16, so those pairs are excluded.

Answer: Before: P(red) = x/18. After: P(red) = (x + 2)/20.
\( \dfrac{x+2}{20} = \dfrac{9}{8} \times \dfrac{x}{18} = \dfrac{x}{16} \)
16x + 32 = 20x, so x = 8.

Teacher's Note:
1. Write the probability before and after adding the balls.
2. Check: 10/20 = 1/2 and 9/8 × 8/18 = 1/2.

Answer: (b) 7. cos θ = 7/25, so tan θ + sec θ = 24/7 + 25/7 = 49/7 = 7.

Teacher's Note:
1. Find cos θ from sin²θ + cos²θ = 1.
2. Then add the two ratios.

Answer: (d) 33. Mode = 3 median − 2 mean gives 45 = 3M − 54, so M = 33.

Teacher's Note:
1. Use the empirical relation.
2. Check: 3 × 33 − 2 × 27 = 45.

Answer: (b) 12 cm. OX ⊥ XY, so XY² = 13² − 5² = 144 and XY = 12.

Teacher's Note:
1. The radius is perpendicular to the tangent.
2. 5, 12, 13 is a Pythagorean triplet.

Answer: (a) 60°. tan θ = 6/(2√3) = √3, so θ = 60°.

Teacher's Note:
1. tan θ is height divided by shadow length.
2. tan 60° = √3.

Answer: (a) 3/26. A deck has 6 red face cards (J, Q, K in hearts and diamonds), so P = 6/52 = 3/26.

Teacher's Note:
1. Face cards are the jack, queen and king of each suit.
2. Only hearts and diamonds are red.

Answer: (b) (3, 0). For (x, 0): (x − 7)² + 36 = (x + 3)² + 16 gives −14x + 85 = 6x + 25, so x = 3.

Teacher's Note:
1. Take the point as (x, 0) and equate squared distances.
2. Check: both distances are √52.

Answer: (d) 25 cm. r² = 24² + 7² = 625, so r = 25.

Teacher's Note:
1. Equate the areas and cancel π.
2. 24, 7, 25 is a Pythagorean triplet.

Answer: (b) 220 cm. 22 km = 2200000 cm, and 2200000 ÷ 10000 = 220 cm.

Teacher's Note:
1. Convert kilometres to centimetres first.
2. Distance = revolutions × circumference.