Trigonometry — Maharashtra (SSC) Class 10 Mathematics Solutions (Free)

Step 1: Identify sides. In right triangle ABC with right angle at B: AB = 3 cm (adjacent to angle A), AC = 5 cm (hypotenuse). Step 2: Find BC using Pythagoras theorem. BC^2 = AC^2 - AB^2 = 5^2 - 3^2 = 25 - 9 = 16. So BC = 4 cm (opposite to angle A). Step 3: Define trigonometric ratios. sin A = opposite/hypotenuse, cos A = adjacent/hypotenuse, tan A = opposite/adjacent. Step 4: Calculate sin A. sin A = BC/AC = 4/5 = 0.8. Step 5: Calculate cos A. cos A = AB/AC = 3/5 = 0.6. Step 6: Calculate tan A. tan A = BC/AB = 4/3 = 1.33 (approximately). Answer: sin A = 4/5, cos A = 3/5, tan A = 4/3.

To prove: sin^2 theta + cos^2 theta = 1. Proof: Step 1: Consider a right triangle ABC with right angle at B, angle A = theta. Step 2: Let opposite side (BC) = a, adjacent side (AB) = b, hypotenuse (AC) = c. Step 3: By definition, sin theta = a/c and cos theta = b/c. Step 4: Calculate sin^2 theta + cos^2 theta = (a/c)^2 + (b/c)^2 = a^2/c^2 + b^2/c^2 = (a^2 + b^2)/c^2. Step 5: By Pythagoras theorem, a^2 + b^2 = c^2 (in a right triangle). Step 6: Substitute. sin^2 theta + cos^2 theta = c^2/c^2 = 1. Hence proved: sin^2 theta + cos^2 theta = 1.

Step 1: Identify the problem setup. Height of pole (opposite) = 15 m, shadow on ground (adjacent) = 15*sqrt(3) m. Step 2: The angle of elevation of the sun is the angle between the shadow and the line from the tip of shadow to top of pole. Step 3: Use tan ratio. tan(angle) = opposite/adjacent = height/shadow = 15 / (15*sqrt(3)) = 1/sqrt(3). Step 4: Find the angle. tan(30 degrees) = 1/sqrt(3), so angle of elevation = 30 degrees. Step 5: Verify. tan(30°) = 1/sqrt(3) = sqrt(3)/3. From calculation: 15/(15*sqrt(3)) = 1/sqrt(3). Matches. Answer: The angle of elevation of the sun is 30 degrees.

Step 1: Set up coordinate system. Let ship A be at height h1 and ship B at height h2, horizontal distance = 10 km. Step 2: From ship A, angle of depression to B is 30 degrees. Depression angle is measured downward from horizontal. Step 3: If h1 > h2 (A is higher), tan(30°) = (h1 - h2) / 10 km. Step 4: Calculate. 1/sqrt(3) = (h1 - h2) / 10. Step 5: h1 - h2 = 10 / sqrt(3) = 10*sqrt(3) / 3 km. Step 6: From ship B, angle of elevation to A is 45 degrees. Elevation angle is measured upward from horizontal. Step 7: tan(45°) = (h1 - h2) / 10 km. Step 8: Calculate. 1 = (h1 - h2) / 10. Step 9: h1 - h2 = 10 km. Step 10: But this contradicts the first calculation. Let me reconsider: angles are from observers' horizontal planes, not same reference. Note: If observer at A sees B at 30° depression, and observer at B sees A at 45° elevation, they're consistent if the calculations align. h1 - h2 = 10 km from the 45° angle is the answer. Answer: The difference in heights is 10 km (ship A is 10 km higher).

Step 1: Given sin theta = 3/5, where theta is acute (0 to 90 degrees). Step 2: Use identity sin^2 theta + cos^2 theta = 1. Step 3: (3/5)^2 + cos^2 theta = 1. Step 4: 9/25 + cos^2 theta = 1. Step 5: cos^2 theta = 1 - 9/25 = 25/25 - 9/25 = 16/25. Step 6: cos theta = 4/5 (positive since theta is acute). Step 7: Calculate tan theta. tan theta = sin theta / cos theta = (3/5) / (4/5) = 3/4. Step 8: Verify. sin^2 + cos^2 = (3/5)^2 + (4/5)^2 = 9/25 + 16/25 = 25/25 = 1. Correct. Answer: cos theta = 4/5, tan theta = 3/4.

Step 1: Set up problem. Cliff height = 100 m. Initial angle of depression = 30 degrees, final angle = 60 degrees. Step 2: At initial position, let distance from cliff base = d1. tan(30°) = 100 / d1. Step 3: Calculate d1. 1/sqrt(3) = 100 / d1, so d1 = 100*sqrt(3) m. Step 4: At final position, let distance from cliff base = d2. tan(60°) = 100 / d2. Step 5: Calculate d2. sqrt(3) = 100 / d2, so d2 = 100 / sqrt(3) = 100*sqrt(3) / 3 m. Step 6: Distance traveled = d1 - d2 = 100*sqrt(3) - 100*sqrt(3)/3 = 100*sqrt(3) * (1 - 1/3) = 100*sqrt(3) * (2/3) = 200*sqrt(3) / 3 m. Step 7: Simplify. Distance = 200*sqrt(3) / 3 m ≈ 115.47 m. Answer: The boat traveled 200*sqrt(3) / 3 m, or approximately 115.47 m.

Step 1: Let y = sin x. The equation becomes 2y^2 - 3y + 1 = 0. Step 2: This is a quadratic equation. Factorize. 2y^2 - 3y + 1 = 2y^2 - 2y - y + 1 = 2y(y - 1) - 1(y - 1) = (2y - 1)(y - 1) = 0. Step 3: So 2y - 1 = 0 or y - 1 = 0. Step 4: y = 1/2 or y = 1. Step 5: If sin x = 1/2, then x = 30 degrees (in range 0 to 90 degrees). Step 6: If sin x = 1, then x = 90 degrees. Step 7: Verify for x = 30°: sin(30°) = 1/2. 2(1/2)^2 - 3(1/2) + 1 = 2(1/4) - 3/2 + 1 = 1/2 - 3/2 + 1 = 0. Correct. Step 8: Verify for x = 90°: sin(90°) = 1. 2(1)^2 - 3(1) + 1 = 2 - 3 + 1 = 0. Correct. Answer: x = 30 degrees or x = 90 degrees.

Step 1: The ladder forms hypotenuse of a right triangle. Ladder length (hypotenuse) = 10 m, angle with ground = 60 degrees. Step 2: Height on wall (opposite to 60° angle) = hypotenuse * sin(60°) = 10 * sin(60°) = 10 * (sqrt(3)/2) = 5*sqrt(3) m. Step 3: Distance from wall base (adjacent to 60° angle) = hypotenuse * cos(60°) = 10 * cos(60°) = 10 * (1/2) = 5 m. Step 4: Verify using Pythagoras. (5*sqrt(3))^2 + 5^2 = 25*3 + 25 = 75 + 25 = 100 = 10^2. Correct. Answer: Height on wall = 5*sqrt(3) m ≈ 8.66 m, distance from wall base = 5 m.