Join 60,000+ competitive exam aspirants
or with email
Your Cart
Subtotal₹0
Total₹0
Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
Join 60,000+ competitive exam aspirants
If A=7.5°, what is the value of (12.8×31cot4A+7.2tan6A)?
12
8
20
10
20
Evaluate the angles first: 4A = 30° and 6A = 45°. Substitute standard values cot 30° = sqrt(3) and tan 45° = 1 to simplify the expression immediately.
A = 7.5°, Expression = (12.8 × (1/sqrt(3)) * cot(4A) + 7.2 * tan(6A))
4A=4×7.5°=30°, 6A=6×7.5°=45°
Evaluate the angles first: 4A = 30° and 6A = 45°. Substitute standard values cot 30° = sqrt(3) and tan 45° = 1 to simplify the expression immediately.
Confusing cot 30° with 1/sqrt(3) or forgetting to multiply A by 4 and 6 before evaluating the trigonometric functions.
Calculate the angles 4A and 6A
Given A=7.5°. Therefore, 4A=4×7.5°=30° and 6A=6×7.5°=45°.
4A=30°,6A=45°
Evaluate trigonometric functions
Substitute the angles into the trigonometric functions: cot(30°)=3 and tan(45°)=1.
cot(30°)=3,tan(45°)=1
Substitute values into the expression
The expression is (12.8×31cot(30°)+7.2tan(45°)). Substitute the values: 12.8×31×3+7.2×1.
12.8×(1)+7.2=12.8+7.2=20
C is correct because the expression simplifies to 12.8 + 7.2, which equals 20.
Always convert the composite angles (like 4A or 6A) to their degree values first; these often result in standard angles like 30°, 45°, or 60° in competitive exams.
Get instant step-by-step AI explanations, full shift Previous Year Papers (SSC GD, RRB, SSC JE), and regional rank analysis on the official Examoogle App.
Free Download • 100K+ Aspirants