CBSE Class 10 Mathematics Trigonometry and Coordinate Geometry Worksheet Set 07

Answer: (a) 17/12. (sin A + cos A) sec A = tan A + 1 = 5/12 + 1 = 17/12.

Teacher's Note:
1. Expand using sec A = 1/cos A.
2. Then substitute tan A.

Answer: (d) 30°. sin θ = 5/10 = 1/2, so θ = 30°.

Teacher's Note:
1. The ladder is the hypotenuse of a right triangle.
2. Height over length gives sin θ.

Answer: (b) 45°. tan θ = 10/10 = 1, so θ = 45°.

Teacher's Note:
1. Height and shadow are equal.
2. tan 45° = 1.

Answer: Let the pole break at height h, so the broken part has length 9 − h. In the right triangle, sin 30° = h/(9 − h) = 1/2, so 2h = 9 − h and h = 3 m.

Teacher's Note:
1. The broken part is the hypotenuse of the triangle.
2. Use sin 30° = 1/2.

Answer: Distance = \( \sqrt{(x-0)^2 + (y-0)^2} = \) √(x² + y²).

Teacher's Note:
1. Use the distance formula with the origin (0, 0).
2. The result is a standard formula.

Answer: Substitute x = 3, y = a: 6 − 3a = 5, so 3a = 1 and a = 1/3.

Teacher's Note:
1. A point lies on a line when its coordinates satisfy the equation.
2. Substitute and solve for a.

Answer: The distance from the x-axis is the absolute value of the y-coordinate, which is 4 units.

Teacher's Note:
1. The distance from the x-axis equals |y|.
2. The distance from the y-axis would be |x| = 3.

Answer: With BC = 3k, AC = 4k, AB = 5k: sin A = 3/5, cos A = 4/5, sin B = 4/5, cos B = 3/5.
(i) \( \tfrac{3}{5}\cdot\tfrac{4}{5} + \tfrac{4}{5}\cdot\tfrac{3}{5} = \tfrac{24}{25} \).
(ii) \( \tfrac{3}{5}\cdot\tfrac{3}{5} + \tfrac{4}{5}\cdot\tfrac{4}{5} = 1 \).
Answers: (i) 24/25 (ii) 1.

Teacher's Note:
1. Draw the triangle with sides 3k, 4k, 5k.
2. Angle B is the complement of angle A.

Answer: Write everything in sin and cos: LHS = \( \dfrac{\cos\theta+1-\sin\theta}{\cos\theta+1+\sin\theta} \).
Multiply the numerator and denominator by (1 + cos θ − sin θ):
Denominator = (1 + cos θ)² − sin²θ = 2cos θ(1 + cos θ).
Numerator = (1 + cos θ − sin θ)² = 2(1 + cos θ)(1 − sin θ).
LHS = \( \dfrac{2(1+\cos\theta)(1-\sin\theta)}{2\cos\theta(1+\cos\theta)} = \dfrac{1-\sin\theta}{\cos\theta} \) = RHS. Hence proved.

Teacher's Note:
1. Convert sec and tan into sin and cos first.
2. Multiply by a suitable conjugate-type term to simplify.

Answer: Let the tower height be h. Initial distance = h cot 30° = h√3; after 6 s the distance = h cot 60° = h/√3.
Distance covered in 6 s = h√3 − h/√3 = 2h/√3.
Speed = (2h/√3)/6 = h/(3√3). Time to cover the remaining h/√3 = (h/√3) ÷ (h/(3√3)) = 3 seconds.

Teacher's Note:
1. Express both car distances in terms of the tower height h.
2. Speed is constant, so use distance divided by speed.

Answer: Let the cloud be at height h above the water and d the horizontal distance from the boy.
The cloud is (h − 5) m above his eyes and its image is (h + 5) m below them.
tan 30° = (h − 5)/d and tan 60° = (h + 5)/d. Dividing: (h + 5)/(h − 5) = tan 60°/tan 30° = 3.
h + 5 = 3h − 15, so h = 10 m.

Teacher's Note:
1. The image is as far below the water as the cloud is above it.
2. Divide the two tangent equations to eliminate d.

Answer: cos 58° = sin 32°, so the first term is 1. sin 24° = cos 66°, so the second term is 1. cos 38° csc 52° = cos 38°/sin 52° = 1.
tan 18° tan 72° = 1 and tan 35° tan 55° = 1, so the denominator is tan 60° = √3.
Value = 1 + 1 − 1/√3 = 2 − 1/√3 = (6 − √3)/3.

Teacher's Note:
1. Use complementary angle identities such as cos(90° − θ) = sin θ.
2. Pair the tangents of complementary angles to get 1.

Answer: Let the centre be (x, y), equidistant from all three points.
From (6, −6) and (3, −7): −12x + 12y + 72 = −6x + 14y + 58, which gives 3x + y = 7.
From (3, −7) and (3, 3): 14y + 49 = −6y + 9, so y = −2 and x = 3.
Centre: (3, −2). Check: its distance from each point is 5.

Teacher's Note:
1. The centre of the circle is equidistant from the three given points.
2. Form two equations by equating squared distances.

Answer: Let the ratio be k : 1. The dividing point is ((3k + 2)/(k + 1), (7k − 2)/(k + 1)).
It lies on the line: 2(3k + 2) + (7k − 2) − 4(k + 1) = 0, so 9k − 2 = 0 and k = 2/9.
The ratio is 2 : 9.

Teacher's Note:
1. Use the section formula with the ratio k : 1.
2. The point of division must satisfy the line's equation.

Answer: The centroid of a triangle coincides with the centroid of the triangle formed by the mid-points of its sides.
Centroid = ((2 + 3 + 6)/3, (1 − 5 + 4)/3) = (11/3, 0).

Teacher's Note:
1. The medial triangle has the same centroid as the original triangle.
2. Average the three coordinates.

Answer: Collinear points give zero area: −6(k + 8) + (−4)(−8 − 10) + 3(10 − k) = 0, so −9k + 54 = 0 and k = 6.
Let B divide AC in m : n. x-coordinate: (3m − 6n)/(m + n) = −4, so 7m = 2n and the ratio is 2 : 7.
AC = √(9² + 18²) = √405 = 9√5 units.

Teacher's Note:
1. Set the area of the triangle formed by the three points equal to zero.
2. Use the section formula for the ratio.

Answer: Area = ½|t[(t + 2) − t] + (t + 2)[t − (t − 2)] + (t + 3)[(t − 2) − (t + 2)]|
= ½|2t + 2(t + 2) − 4(t + 3)| = ½|2t + 2t + 4 − 4t − 12| = ½|−8| = 4 sq. units.
The area does not contain t, so it is independent of t.

Teacher's Note:
1. Use the area formula for three vertices.
2. Simplify until t cancels out.

Answer: Let the ratio be k : 1. The y-coordinate gives (2k − 3)/(k + 1) = −1, so 3k = 2 and k = 2/3. The ratio is 2 : 3.
p = (6k + 1)/(k + 1) = (4 + 1)/(5/3) = 3.

Teacher's Note:
1. Use the known y-coordinate −1 to find the ratio.
2. Then use the ratio to find the x-coordinate.

Answer: AB = √26, BC = √26, CD = √26, DA = √26, so all sides are equal and ABCD is a rhombus.
Diagonals: AC = √(16 + 16) = 4√2 and BD = √(36 + 36) = 6√2. They are unequal, so it is not a square.
Area = ½ × AC × BD = ½ × 4√2 × 6√2 = 24 sq. units.

Teacher's Note:
1. Find all four sides and both diagonals with the distance formula.
2. A rhombus is a square only when its diagonals are equal.