Trigonometry (त्रिकोणमिती) MCQ Test — Class 9 Maharashtra SSC

Solution:

Identify the sides relative to angle A. The side opposite to angle A is BC = 6 cm. The side adjacent to angle A is AB = 8 cm.

Apply the definition of tan A. — tan A = Opposite / Adjacent

Substitute the given values. — tan A = BC / AB = 6 / 8

Simplify the fraction. — tan A = 3 / 4

Solution:

For an acute angle A, sin A and cos A are ratios of a leg to the hypotenuse. Since the hypotenuse is always the longest side, sin A and cos A are always less than or equal to 1.

sec A is the reciprocal of cos A. Since cos A ≤ 1, sec A (1/cos A) must be greater than or equal to 1 for an acute angle.

tan A is the ratio of the opposite side to the adjacent side. In a right-angled triangle, the opposite side can be longer than the adjacent side, or vice-versa. Therefore, tan A can be less than 1, equal to 1 (for 45°), or greater than 1.

cos A = Adjacent / Hypotenuse. Since Adjacent < Hypotenuse, cos A must always be less than 1. So, 5/3 (which is > 1) is not possible for cos A.

Solution:

The cosecant of an angle is the reciprocal of its sine.

Write down the reciprocal relationship. — cosec θ = 1 / sin θ

Substitute the given value of sin θ. — cosec θ = 1 / (12/13)

Simplify the expression. — cosec θ = 13 / 12

Solution:

Recall the standard values: sin 30° = 1/2 and tan 45° = 1.

Substitute these values into the expression. — 2 × sin 30° + tan 45° = 2 × (1/2) + 1

Perform the multiplication. — = 1 + 1

Perform the addition. — = 2

Solution:

In a right-angled triangle PQR, right-angled at Q, PR is the hypotenuse.

For angle P, the side adjacent to it is PQ.

The definition of cos P is Adjacent / Hypotenuse.

Therefore, cos P = PQ / PR. Ravi's formula is correct.

Solution:

Recall the values: sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2.

Convert to approximate decimals for easier comparison: sin 30° = 0.5, sin 45° ≈ 0.707, sin 60° ≈ 0.866.

Compare the values: 0.5 < 0.707 < 0.866. This means sin 30° < sin 45° < sin 60°.

Based on this, sin 60° > sin 45° is the correct statement.

Solution:

Use the identity sin²A + cos²A = 1.

Substitute the given value of sin A. — (3/5)² + cos²A = 1

Calculate the square of sin A. — 9/25 + cos²A = 1

Isolate cos²A and solve for cos A. — cos²A = 1 - 9/25 = 16/25

Take the square root. Since A is an acute angle, cos A is positive. — cos A = √(16/25) = 4/5

Solution:

Form a right-angled triangle. The ladder is the hypotenuse, the distance from the wall is the adjacent side to the 60° angle, and the wall forms the opposite side.

We know the adjacent side (2 m) and the angle (60°), and we need to find the hypotenuse (length of the ladder). The cosine ratio relates these three.

Apply the cosine formula. — cos(angle) = Adjacent / Hypotenuse

Substitute the values and solve for the hypotenuse (ladder length 'L'). — cos 60° = 2 / L

Recall cos 60° = 1/2. So, 1/2 = 2 / L. This gives L = 2 × 2 = 4 m.

Solution:

The fundamental trigonometric ratios are defined based on the sides of a right-angled triangle relative to an acute angle.

SOH: Sine = Opposite / Hypotenuse.

CAH: Cosine = Adjacent / Hypotenuse.

TOA: Tangent = Opposite / Adjacent.

Therefore, 'Adjacent side / Hypotenuse' represents cos A.

Solution:

In triangle PQR, right-angled at Q, PR is the hypotenuse.

For angle R: the side opposite is PQ, and the side adjacent is QR.

Recall that sec R is the reciprocal of cos R. So, sec R = 1 / cos R.

First, find cos R. cos R = Adjacent / Hypotenuse = QR / PR.

Therefore, sec R = 1 / (QR / PR) = PR / QR.