Introduction to Trigonometry — Telangana (SSC) Class 10 Mathematics Solutions (Free)
Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Introduction to Trigonometry" — important questions with detailed answers, download PDF for board exam preparation.
TL;DR: Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Introduction to Trigonometry" — important questions with detailed answers, downl…
By Syllab.in · Updated
Q1: Define the six trigonometric ratios for a right triangle. State the trigonometric values for the standard angles 0°, 30°, 45°, 60°, and 90°.
Trigonometric Ratios (for a right triangle with reference angle θ):
Let a right triangle have:
- Opposite side (opposite to angle θ) = opposite
- Adjacent side (adjacent to angle θ) = adjacent
- Hypotenuse = hypotenuse
The six ratios are:
1. sin θ = opposite / hypotenuse
2. cos θ = adjacent / hypotenuse
3. tan θ = opposite / adjacent
4. cot θ = adjacent / opposite = 1/tan θ
5. sec θ = hypotenuse / adjacent = 1/cos θ
6. cosec θ = hypotenuse / opposite = 1/sin θ
MNEMONIC: SOH-CAH-TOA
Sin=Opposi…
Q2: In a right triangle ABC with right angle at B, AB = 5 cm and AC = 13 cm. Find all six trigonometric ratios of angle A.
Given:
Right triangle ABC with ∠B = 90°
AB = 5 cm (side adjacent to angle A)
AC = 13 cm (hypotenuse)
Find: All six trigonometric ratios of angle A
Step 1: Find BC (the side opposite to angle A) using Pythagorean theorem.
AC² = AB² + BC²
13² = 5² + BC²
169 = 25 + BC²
BC² = 144
BC = 12 cm
Step 2: Identify the sides relative to angle A.
Opposite side (opposite to ∠A) = BC = 12 cm
Adjacent side (adjacent to ∠A) = AB = 5 cm
Hypotenuse = AC = 13 cm
Step 3: Calculate sin A.
sin A = opposite / hypot…
Q3: Prove the identity: sin²θ + cos²θ = 1
Prove: sin²θ + cos²θ = 1
Proof:
Consider a right triangle with:
- Angle θ at one vertex
- Opposite side = a
- Adjacent side = b
- Hypotenuse = c
Step 1: Write the definitions of sin θ and cos θ.
sin θ = a/c
cos θ = b/c
Step 2: Square both ratios.
sin²θ = a²/c²
cos²θ = b²/c²
Step 3: Add the squared ratios.
sin²θ + cos²θ = a²/c² + b²/c²
sin²θ + cos²θ = (a² + b²)/c²
Step 4: Apply the Pythagorean theorem.
In a right triangle: a² + b² = c²
Substitute:
sin²θ + cos²θ = c²/c²
sin²θ + cos²θ = 1
He…
Q4: Simplify: (sin 30° + cos 60°)/(tan 45°) + sec 60° - cosec 30°
Given: Simplify (sin 30° + cos 60°)/(tan 45°) + sec 60° - cosec 30°
Step 1: Recall the values of trigonometric functions for standard angles.
sin 30° = 1/2
cos 60° = 1/2
tan 45° = 1
sec 60° = 2
cosec 30° = 2
Step 2: Substitute these values.
= (1/2 + 1/2)/(1) + 2 - 2
Step 3: Simplify the first term.
= (1)/(1) + 2 - 2
= 1 + 2 - 2
= 1
Answer: The simplified expression equals 1
Q5: If sin θ = 3/5 and θ is acute, find cos θ and tan θ.
Given: sin θ = 3/5, θ is acute (0° < θ < 90°)
Find: cos θ and tan θ
Step 1: Use the Pythagorean identity sin²θ + cos²θ = 1.
sin²θ + cos²θ = 1
(3/5)² + cos²θ = 1
9/25 + cos²θ = 1
cos²θ = 1 - 9/25
cos²θ = 25/25 - 9/25
cos²θ = 16/25
Step 2: Take the square root.
cos θ = ±4/5
Since θ is acute, cos θ > 0.
cos θ = 4/5
Step 3: Find tan θ.
tan θ = sin θ / cos θ
tan θ = (3/5) / (4/5)
tan θ = (3/5) × (5/4)
tan θ = 3/4
Answer:
cos θ = 4/5 = 0.8
tan θ = 3/4 = 0.75
Verification:
sin²θ + cos²θ…
Showing 5 of 9 questions — full solutions on the page.