Trigonometry — Maharashtra (SSC) Class 10 Mathematics Solutions (Free)
Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Trigonometry" — important questions with detailed answers, download PDF for board exam preparation.
TL;DR: Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Trigonometry" — important questions with detailed answers, download PDF for bo…
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Q1: In a right triangle ABC with right angle at B, AB = 3 cm and AC = 5 cm. Find sin A, cos A, and tan A.
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.…
Q2: Prove the identity sin^2 theta + cos^2 theta = 1.
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.…
Q3: A pole of height 15 m casts a shadow of 15*sqrt(3) m on the ground. Find the angle of elevation of the sun.
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: …
Q4: Two ships are 10 km apart. From ship A, ship B is observed at angle 30 degrees below horizontal. From ship B, ship A is at angle 45 degrees above horizontal. Find the difference in heights of the two ships.
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 hor…
Q5: If sin theta = 3/5 and theta is acute, find cos theta and tan theta.
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, …
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