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Xcelerate 3.0 (Ideathon)
Table of content:
- Number System: Important Divisibility Rules
- LCM & HCF: Important Formulas
- Simplification & Approximation Formulas
- Percentage: Basics & Important Formulas
- Profit, Loss & Discount: Important Formulas
- Ratio & Proportion: Important Formulas
- Simple & Compound Interest: Important Formulas
- Time, Speed & Distance: Important Formulas
- Time and Work: Pipes & Cisterns Formulas
- Average: Key Concepts & Formula
- Mixtures & Alligation Formulas
- Basic Geometry Formulas
- Important Trigonometric Formulas
- Conclusion
- Frequently Asked Questions (FAQs)
Best Cheat Sheet For Quantitative Aptitude: Top 13 Formulas Made Easy
A cheat sheet for quantitative aptitude formulas will boost your speed, accuracy, and confidence in government or company placement exams! Read on to get the details.
17 mins read
This cheat sheet covers all essential formulas, shortcuts, and tricks to help you solve problems faster and more accurately. These formulas will boost your speed, accuracy, and confidence. Practice smart and excel in your exams.
Number System: Important Divisibility Rules
Here is a proper table format for the Divisibility Rules:
| Divisibility by | Rule | Example |
|---|---|---|
| 1 | All numbers are divisible by 1. | Any number |
| 2 | If the last digit is even (0, 2, 4, 6, 8), the number is divisible by 2. | 128 Ends in 8 |
| 3 | If the sum of the digits is divisible by 3, the number is divisible by 3. | 273 (2+7+3) = 12, 12 ÷ 3 = 4 |
| 4 | If the last two digits form a number divisible by 4, the number is divisible by 4. | 712 Last two digits = 12 → 12 ÷ 4 = 3 |
| 5 | If the number ends in 0 or 5, it is divisible by 5. | 675 Ends in 5 |
| 6 | If the number is divisible by both 2 and 3, it is divisible by 6. | 108 Divisible by 2 (ends in 8) & by 3 (sum = 9) |
| 7 | Double the last digit, subtract from the rest of the number. If the result is divisible by 7, so is the number. | 203 (3 × 2 = 6), 20 - 6 = 14 (divisible by 7) |
| 8 | If the last three digits form a number divisible by 8, then the whole number is divisible by 8. | 816 816 ÷ 8 = 102 |
| 9 | If the sum of the digits is divisible by 9, the number is divisible by 9. | 729 (7+2+9) = 18, and 18 ÷ 9 = 2 |
| 10 | If the number ends in 0, it is divisible by 10. | 340 Ends in 0 |
| 11 | If the difference between the sum of digits in odd places and even places is divisible by 11, the number is too. | 121 (1+1) - (2) = 0 (divisible by 11) |
| 12 | If the number is divisible by both 3 and 4, it is divisible by 12. | 864 Divisible by 3 (sum = 18), by 4 (last two digits 64 ÷ 4 = 16) |
LCM & HCF: Important Formulas
| Concept | Formula / Rule | Example |
|---|---|---|
| 1. LCM & HCF Relationship | LCM × HCF = Product of Numbers (for two numbers) |
LCM(12,18) = 36 HCF(12,18) = 6 6×36=12×18=216 |
| 2. HCF using Prime Factorization | Take the smallest power of common prime factors. |
HCF(24, 36): 36 = 22×32 HCF = 22×31=12 |
| 3. LCM using Prime Factorization | Take the highest power of all prime factors. |
LCM(24, 36): 36 = 22×32 |
| 4. HCF using Division (Euclidean Algorithm) |
HCF(a,b)=HCF(b,amod b) (repeat until remainder = 0) |
HCF(56, 98): 98 ÷ 56 remainder 42 56 ÷ 42 remainder 14 42 ÷ 14 remainder 0 HCF = 14 |
| 5. LCM of Fractions | LCM = LCM of Numerators ÷ HCF of Denominators | LCM of 2÷3,5÷6 LCM(2,5) = 10, HCF(3,6) = 3 LCM = 10/3 |
| 6. HCF of Fractions | HCF = HCF of Numerators ÷ LCM of Denominators | HCF of 4÷9,10÷15 HCF(4,10) = 2, LCM(9,15) = 45 HCF = 2/45 |
Simplification & Approximation Formulas
| Concept | Formula / Rule | Example |
|---|---|---|
| BODMAS Rule | Brackets → Orders (powers/roots) → Division → Multiplication → Addition → Subtraction | 8+2×5−32 = 8+10−9 = 9 |
| Multiplication Short Trick | (a + b) × (a - b) = a² - b² | 102×98 = (100+2)(100−2)=10000−4=9996 |
| Square of a Number Ending in 5 | n2=(n−1)×(n+1)+25 | 352=(30×40)+25=1225 |
| Percentage to Fraction Conversion | x%=x/100 | 25%=25/100=1/4 |
| Fraction to Percentage Conversion | x/y×100% | 3/5=3/5×100=60% |
| Average Formula |
Average=Sum of all values ÷ Number of values |
Avg of 5,10,15=(5+10+15)/3=10 |
| Compound Interest Approximation | A=P(1+r/100)t |
If P=1000,r=10%,t=2 A=1000(1.1)2≈1210 |
| Speed, Distance & Time | Speed=Distance÷Time | If distance = 150 km, time = 3 hr → Speed = 150/3 = 50 km/hr |
| Approximate Square Root | Use nearest perfect squares | √50≈√49=7 |
| Ratio & Proportion | a:b=c:d⇒ad=bc | If 2:3=4:x, then 2x=12⇒x=6 |
Percentage: Basics & Important Formulas
| Concept | Formula / Rule | Example |
|---|---|---|
| Percentage Definition | x%=x/100 | 25%=25/100=0.25 |
| Conversion: Fraction ↔ Percentage | x/y×100% | 3/5=(3/5)×100=60% |
| Percentage Increase | New Value=Old Value×(1+Increase % ÷ 100) | If price = ₹500, increase = 10% New Price = 500×1.1=₹550 |
| Percentage Decrease | New Value=Old Value×(1−Decrease % ÷ 100) | If price = ₹400, decrease = 20% New Price = 400×0.8 |
| Percentage Change Formula | Change in Value ÷ Original Value×100 | Increase from 50 to 65: 65−50÷ 50×100=30% |
| Finding What % is A of B | A/B×100 | What % is 20 of 50? (20/50)×100=40% |
| Finding A When B% is Given | A=B×Given % ÷ 100 | Find 25% of 200: 200×(25/100)=50 |
| Successive Percentage Change | Final Change=a+b+ab ÷ 100 | Increase 10%, then 20%: 10+20+10×20 ÷ 100=32% |
| Population Growth/Depreciation | P=P0×(1±r÷100)t | Initial pop = 5000, growth = 5% per year for 2 years 5000×(1.05)2=5525 |
| Profit & Loss Percentage | Profit %=Profit ÷ Cost Price×100 Loss %=Loss ÷ Cost Price×100 |
CP = ₹200, SP = ₹250 Profit% = (50/200)×100=25% |
Profit, Loss & Discount: Important Formulas
| Concept | Formula / Rule | Example |
|---|---|---|
| 1. Profit Formula | Profit = Selling Price (SP) - Cost Price (CP) | SP = ₹250, CP = ₹200 → Profit = ₹250 - ₹200 = ₹50 |
| 2. Loss Formula | Loss = Cost Price (CP) - Selling Price (SP) | SP = ₹150, CP = ₹200 → Loss = ₹200 - ₹150 = ₹50 |
| 3. Profit Percentage | Profit % = Profit ÷ Cost Price×100 | Profit = ₹50, CP = ₹200 → Profit % = 50÷ 200×100=25% |
| 4. Loss Percentage | Loss % = Loss÷ Cost Price×100 | Loss = ₹50, CP = ₹200 → Loss % = 50÷ 200×100=25% |
| 5. Cost Price from Profit Percentage | Cost Price = Selling Price ÷ 1+Profit %/100 | SP = ₹250, Profit % = 25 → CP = 250÷1+0.25=₹200 |
| 6. Selling Price from Profit Percentage | CP = ₹200, Profit % = 25 → SP = 200×1.25=₹250 | |
| 7. Marked Price (MP) & Discount Formula | Discount = Marked Price (MP) - Selling Price (SP) | MP = ₹500, SP = ₹400 → Discount = ₹500 - ₹400 = ₹100 |
| 8. Discount Percentage | Discount % = Discount÷ Marked Price×100 | Discount = ₹100, MP = ₹500 → Discount % = 100÷ 500×100=20% |
| 9. Selling Price with Discount | Selling Price = Marked Price ×(1−Discount %÷ 100) | MP = ₹500, Discount % = 20 → SP = 500×0.8=₹400 |
| 10. Cost Price with Profit Percentage | Cost Price = Selling Price÷ 1+Profit % ÷ 100 | SP = ₹500, Profit % = 20 → CP = 500 ÷ 1+0.2=₹416.67 |
Ratio & Proportion: Important Formulas
| Concept | Formula / Rule | Example |
|---|---|---|
| 1. Basic Ratio | a/b=c | If a:b = 2:3, then a/b=2/3 |
| 2. Simplifying Ratios | Divide both terms by their HCF | For ratio 12:16, HCF = 4 → Simplified ratio = 3:4 |
| 3. Ratio of Quantities | Ratio=Quantity 1 ÷ Quantity 2 | If quantity 1 = 30, quantity 2 = 50 → Ratio = 30÷50=3:5 |
| 4. Proportion Formula | a/b=c/d | If 2/4=3/6, then they are in proportion |
| 5. Cross Multiplication | a×d=b×c | For a/b=c/dcross multiply: a×d=b×c |
| 6. Mean Proportional | Mean Proportional=√a×b | For 4 and 9, mean proportional = √4×9=6 |
| 7. Alt. Mean Proportional | a/b=b/c⇒b2=ac | For 3/6=6/12\, we can find the mean proportional by 62=3×12 |
| 8. Direct Proportion | y=kx, where is constant | If y = 3x, and x = 4, then y = 12. |
| 9. Inverse Proportion | y=k/x, where k is constant | If y = 12/x, and x = 4, then y = 3. |
| 10. Proportion Rule for Dividing a Quantity in a Ratio | First part/Total parts=Ratio part/Sum of parts | For dividing ₹5000 in the ratio 2:3, each part = 5000/2+3=1000, so first part = ₹2000, second part = ₹3000. |
Simple & Compound Interest: Important Formulas
| Concept | Formula / Rule | Example |
|---|---|---|
| 1. Simple Interest (SI) | SI = P×R×T/100 |
P = ₹1000, R = 5%, T = 2 years SI = 1000×5×2/100=₹100 |
| 2. Amount (A) in Simple Interest | A = P + SI |
P = ₹1000, SI = ₹100 A = ₹1000 + ₹100 = ₹1100 |
| 3. Compound Interest (CI) | CI = P (1+R/100)T−P |
P = ₹1000, R = 5%, T = 2 years CI = 1000×(1+5/100)2−1000=₹102.50 |
| 4. Amount (A) in Compound Interest | A = P (1+R/100)T |
P = ₹1000, R = 5%, T = 2 years A = 1000×(1+5/100)2=₹1102.50 |
| 5. Simple Interest for a Given Principal | P = SI×100/R×T |
SI = ₹100, R = 5%, T = 2 years P = 100×100/5×2=₹1000 |
| 6. Rate of Interest (R) | R = SI×100/P×T |
SI = ₹100, P = ₹1000, T = 2 years R = 100×100/1000×2=5% |
| 7. Time Period (T) | T = SI×100/P×R |
SI = ₹100, P = ₹1000, R = 5% T = 100×100/1000×5=2 years |
| 8. Compound Interest with Annual Compounding | CI = A - P |
A = ₹1102.50, P = ₹1000 CI = ₹1102.50 - ₹1000 = ₹102.50 |
| 9. Compound Interest with Half-Yearly Compounding | A = P (1+R/200)2T |
P = ₹1000, R = 10%, T = 1 year A = 1000×(1+10/200)2×1=₹1102.50 |
| 10. Compound Interest with Quarterly Compounding | A = P (1+R/400)4T |
P = ₹1000, R = 12%, T = 1 year A = 1000×(1+12/400)4×1=₹1123.61 |
Time, Speed & Distance: Important Formulas
| Concept | Formula / Rule | Example |
|---|---|---|
| 1. Basic Formula | Speed = Distance/Time |
Distance = 100 km, Time = 2 hours Speed = 100/2=50 km/h |
| 2. Time Calculation | Time = Distance/Speed |
Distance = 120 km, Speed = 60 km/h Time = 120/60= 2hrs |
| 3. Distance Calculation | Distance = Speed × Time |
Speed = 60 km/h, Time = 3 hours Distance = 60×3=180 km |
| 4. Relative Speed (Two Objects) | Relative Speed = Speed of A ± Speed of B | Speed of A = 50 km/h, Speed of B = 30 km/h Relative Speed = 50+30=80 km/h (if moving in the same direction) |
| 5. Trains Crossing Each Other | Time = Length of Train 1 + Length of Train 2/Relative Speed |
Length of Train 1 = 120 m, Length of Train 2 = 150 m, Relative Speed = 60 km/h Time = 120+150/60×5/18=30 seconds |
| 6. Train Crossing a Pole or Bridge | Time = Length of Train/Speed |
Length of Train = 100 m, Speed = 72 km/h Time = 100/72×5/18=5 seconds |
| 7. Boats and Streams (Upstream) | Speed (Upstream) = Speed of Boat - Speed of Stream |
Speed of Boat = 10 km/h, Speed of Stream = 2 km/h Speed (Upstream) = 10−2=8 km/h |
| 8. Boats and Streams (Downstream) | Speed (Downstream) = Speed of Boat + Speed of Stream |
Speed of Boat = 10 km/h, Speed of Stream = 2 km/h Speed (Downstream) = 10+2=12 km/h |
| 9. Time in Upstream | Time = Distance/Speed (Upstream) |
Distance = 20 km, Speed (Upstream) = 8 km/h Time = 20/8=2.5 hours |
| 10. Time in Downstream | Time = Distance/Speed (Downstream) |
Distance = 24 km, Speed (Downstream) = 12 km/h Time = 24/12=2 hours |
| 11. Speed of Boat in Still Water | Speed of Boat = Speed (Downstream)+Speed (Upstream)/2 |
Speed (Downstream) = 12 km/h, Speed (Upstream) = 8 km/h Speed of Boat = 12+8/2=10 km/h |
| 12. Speed of Stream | Speed of Stream = Speed (Downstream)−Speed (Upstream)/2 |
Speed (Downstream) = 12 km/h, Speed (Upstream) = 8 km/h Speed of Stream = 12−8/2=2 km/h |
Time and Work: Pipes & Cisterns Formulas
| Concept | Formula / Rule | Example |
|---|---|---|
| 1. Work Done by One Pipe | Work = 1/Time taken by the pipe | If a pipe fills a tank in 4 hours → Work = 1/4 of the tank per hour. |
| 2. Combined Work | Work = 1/Time taken by pipe A+1/Time taken by pipe B | Pipe A fills the tank in 6 hours, Pipe B in 8 hours → Combined Work = 1/6+1/8=7/24 |
| 3. Time Taken to Fill/Empty Tank | Time = 1/Combined Work | If combined work is 7/24, then time = 1/7/24=3.43 hours to fill the tank. |
| 4. Emptying a Tank (Negative Work) | Work = 1/Time taken to empty the tank | If a pipe empties the tank in 10 hours → Work = −1/10 (negative because it empties). |
| 5. Time to Fill or Empty the Tank with Two Pipes | Time = 1/(1/Time taken by Pipe 1±1/Time taken by Pipe 2) | Pipe A fills in 6 hours, Pipe B empties in 8 hours → Time to fill the tank = 1(1/6−1/8) |
| 6. If Pipe A Fills & Pipe B Empties | Time = 1(1/A−1/B) | Pipe A fills in 10 hours, Pipe B empties in 12 hours → Time = 1(1/10−1/12)=60 hours |
| 7. Work Done in x Days (Multiple Pipes) | Work = Number of workers×Time taken by each worker | If 5 pipes can fill a tank in 12 days → Work = 5×12=60 units of work. |
| 8. Pipes Working in Cycles | Time = 1/(1/T1+1/T2+⋯+1/Tn) | If Pipe 1 fills in 6 hours, Pipe 2 in 8 hours → Combined Time = 1(1/6+1/8) |
| 9. Efficiency of Pipes | Efficiency = Work done/Time taken | If a pipe fills a tank in 5 hours → Efficiency = 1/5 of the tank per hour. |
Average: Key Concepts & Formula
| Concept | Formula / Rule |
|---|---|
| 1. Average of n Numbers | Average = Sum of all values/n |
| 2. Weighted Average | Weighted Average = ∑(xi×wi)/∑wi |
| 3. Average Speed (Two Distances) | Average Speed = 2×Speed 1×Speed 2/Speed 1+Speed 2 |
| 4. Average of Grouped Data (Class Interval) | Average = ∑(fi×xi)/∑fi |
| 5. Cumulative Average | Cumulative Average = Sum of first n terms/n |
| 6. Average of Negative Numbers | Same as the Average of Positive Numbers |
| 7. Harmonic Mean | Harmonic Mean = n/∑(1/xi) |
| 8. Arithmetic Mean (AM) | AM = a1+a2+a3+⋯+an/n |
| 9. Median Formula for Odd Numbers | Median = Middle value |
| 10. Median Formula for Even Numbers | Median = (n/2)th term + (n/2 + 1)th term/2 |
Exam Tips on Average
| Tip | Explanation |
|---|---|
| Weighted Average | Use when values have different weights or importance. Multiply each value by its weight, then divide by the sum of weights. |
| Average Speed | When a journey has two parts with different speeds, use the Average Speed formula to calculate the overall speed. |
| Cumulative Average | Keep track of the average as new terms are added by dividing the sum of terms by the total number of terms. |
| Grouped Data | For grouped data, use the midpoint of each class interval and multiply by its frequency to find the average. |
| Harmonic Mean | Use the Harmonic Mean for rates, speeds, and times, especially when the values vary inversely. |
Mixtures & Alligation Formulas
| Concept | Formula / Rule | Example |
|---|---|---|
| 1. Alligation (Mean) | Alligation Mean = (C1×W1)+(C2×W2)/W1+W2 | A mixture of 10 liters of liquid costing ₹20 per liter is mixed with 20 liters of liquid costing ₹30 per liter. What is the mean cost of the mixture? Solution: Mean Cost=(20×10)+(30×20)/10+20= ₹26.67 |
| 2. Alligation Alternate | Alligation Alternate: Difference between the mean and the prices of each component, used to find the required proportions. | A mixture of water and milk has a mean price of ₹40 per liter. If water costs ₹30 and milk costs ₹50 per liter, find the ratio of water and milk in the mixture. Solution: Difference between mean and water price = 40−30=10; Difference between mean and milk price = 50−40=10. Therefore, the ratio of water to milk is 1:1. |
| 3. Alligation Rule (Simple Mixture) | Proportion 1 = C2−M/C2−C1 and Proportion 2 = M−C1/C2−C1 |
A mixture of 40 liters of water costing ₹5 per liter and 60 liters of juice costing ₹8 per liter is made. What is the mean price? |
| 4. Cost of Mixing Two Ingredients | Cost Price of Mixture = (C1×W1)+(C2×W2)/W1+W2 | If 10 kg of sugar costing ₹50 per kg is mixed with 15 kg of sugar costing ₹60 per kg, find the cost price of the mixture. Solution: Cost of Mixture=(50×10)+(60×15)/10+15=500+900/25=56 |
| 5. Mixture of Milk & Water | Proportion of Milk = C2−M/C2−C1 and Proportion of Water = M−C1/C2−C1 |
In a milk-water mixture, the cost of milk is ₹10 per liter and the cost of water is ₹5 per liter. If the mean price of the mixture is ₹8 per liter, find the proportion of milk and water. |
| 6. Quantity of Components in Mixture | Quantity of Component 1 = C2−M/C2−C1×W1 | A mixture of 40 liters of water costing ₹5 per liter and 60 liters of juice costing ₹8 per liter is made. Find the quantity of juice needed for the mixture to have a mean price of ₹6. Solution: Quantity of Juice=(6−5/(8−5)×40=1/3×40=13.33 |
| 7. Alligation for Discounted Mixture | For discount problems, calculate the difference between the selling price and the cost price and use the Alligation Method for determining the required proportions. | A shopkeeper buys two kinds of goods at ₹10 and ₹20 per kg. If he wants to sell the mixture at ₹16 per kg, what should be the proportion of the two types of goods? Solution: 16−10=6 and 20−16=4, so the ratio is 4:6 or 2:3. |
| 8. Mixture Cost with Percentage | Mixture Cost = (C1×W1)+(C2×W2)/W1+W2 | A mixture of two liquids, 10 liters of one costing ₹50 per liter and 15 liters of the other costing ₹60 per liter, is made. What is the cost price of the mixture? Solution: Cost of Mixture=(50×10)+(60×15)/10+15=500+900/25=56 |
Tips for Solving Mixture Problems
| Tip | Explanation |
|---|---|
| Alligation Method | A technique to find the mean price of a mixture by using the prices and quantities of two substances. |
| Alligation Alternate Rule | A shortcut method where the differences between the mean and component prices are calculated and cross-multiplied to find proportions. |
| Milk & Water Problems | Use Alligation to mix two substances like milk and water at different prices to get a desired mean price. |
| Mixture of Two Different Prices | Use Cost of Mixing to calculate the cost of the mixture based on the prices and quantities of the components. |
| Discount Mixtures | For problems involving discounts, use Alligation to figure out the right proportions of different components. |
Basic Geometry Formulas
| Concept | Formula | Explanation | Example |
|---|---|---|---|
| 1. Perimeter of a Rectangle | P=2×(l+b) | Perimeter is the sum of all sides of the rectangle. | For a rectangle with length 5 units and breadth 3 units, P=2×(5+3)=16 units. |
| 2. Perimeter of a Square | P=4×a | A square has four equal sides, so the perimeter is 4 times the length of one side. | For a square with side length 4 units, P=4×4=16. |
| 3. Perimeter of a Triangle | P=a+b+c | The perimeter of a triangle is the sum of the lengths of its three sides. | For a triangle with sides 3, 4, and 5 units, P=3+4+5 units. |
| 4. Area of a Rectangle | A=l×b | Area is the product of length and breadth for a rectangle. | For a rectangle with length 5 units and breadth 3 units, A=5×3= square units. |
| 5. Area of a Square | A=a2 | The area of a square is the square of its side length. | For a square with side length 4 units, A=42= square units. |
| 6. Area of a Triangle | A=1/2×b×h | Area is half the product of the base and the height of the triangle. | For a triangle with base 6 units and height 4 units, A=1/2×6×4=12 square units. |
| 7. Area of a Circle | A=πr2 | The area of a circle is π times the square of the radius. | For a circle with radius 3 units, A=π×32=28.27 square units (approximately). |
| 8. Circumference of a Circle | C=2πr | The circumference (perimeter) of a circle is 2π times the radius. | For a circle with radius 3 units, C=2π×3 units (approximately). |
| 9. Area of a Parallelogram | A=b×h | The area of a parallelogram is the product of its base and height. | For a parallelogram with base 6 units and height 4 units, A=6×4=24 square units. |
| 10. Area of a Rhombus | A=1/2×d1×d2 | The area of a rhombus is half the product of its diagonals. | For a rhombus with diagonals of 5 units and 8 units, A=1/2×5×8=20 square units. |
| 11. Angle Sum of Triangle | Sum of Angles=180° | The sum of the interior angles of a triangle is always 180 degrees. | In any triangle, the sum of all interior angles is 180°. |
| 12. Complementary Angles | A+B=90° | Two angles are complementary if their sum is 90 degrees. | If A=30°, then B=90°−30°=60°. |
| 13. Supplementary Angles | A+B=180° | Two angles are supplementary if their sum is 180 degrees. | If A=110°, then B=180°−110°=70°. |
| 14. Vertical Angles | Vertical Angles are equal | When two lines intersect, opposite angles (vertical angles) are always equal. | If one vertical angle is 50°, the opposite vertical angle is also 50°. |
| 15. Exterior Angle of a Triangle | Exterior Angle=Sum of Opposite Interior Angles | The exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles. | For a triangle with angles 40°, 60°, and an exterior angle of 100°, the exterior angle is equal to 40°+60°=100° |
Important Trigonometric Formulas
For a right-angled triangle with:
- Hypotenuse = h
- Opposite side = p
- Adjacent side = b
| Function | Formula |
|---|---|
| Sine (sin θ) | sinθ=p/h |
| Cosine (cos θ) | cosθ=b/h |
| Tangent (tan θ) | tanθ=p/b |
| Cosecant (csc θ) | cscθ=h/p |
| Secant (sec θ) | secθ=h/b |
| Cotangent (cot θ) | cotθ=b/p |
Values for Trigonometric Angles
| Angle (θ) | sin θ | cos θ | tan θ | cosec θ | sec θ | cot θ |
|---|---|---|---|---|---|---|
| 0° | 0 | 1 | 0 | ∞ | 1 | ∞ |
| 30° | 1/2 | √3/2 | 1/√3 | 2 | 2/√3 | √3 |
| 45° | √2/2 | √2/2 | 1 | √2 | √2 | 1 |
| 60° | √3/2 | 1/2 | √3 | 2/√3 | 2 | 1/√3 |
| 90° | 1 | 0 | ∞ | 1 | ∞ | 0 |
Most Common Trigonometric Identities
| Identity Type | Formula |
|---|---|
| 1. Pythagorean Identities |
sin2θ+cos2θ=1 1+tan2θ=sec2θ 1+cot2θ=csc2θ |
| 2. Reciprocal Identities |
cscθ=1/sinθ secθ=1/cosθ cotθ=1/tanθ |
| 3. Co-function Identities |
sin(90°−θ)=cosθ, cos(90°−θ)=sinθ tan(90°−θ)=cotθ, cot(90°−θ)=tanθ |
| 4. Angle Sum & Difference |
sin(A±B)=sinA cosB ± cosA sinB cos(A±B)=cosA cosB ∓ sinA sinB tan(A±B)=1∓ tanA tanB tanA ± tanB |
| 5. Double Angle Formulas |
sin2A=2sinA cosA cos2A=cos2A−sin2A=2cos2A−1=1−2sin2A tan2A=2tanA/1-tan2A |
| 6. Law of Sines | a/sinA=b/sinB=c/sinC |
| 7. Quotient Identities |
tanθ=sinθ/cosθ cotθ=cosθ/sinθ |
| 8. Law of Cosines | c2=a2+b2−2ab cosC |
Angle of Elevation & Angle of Depression
| Concept | Definition | Key Formula |
|---|---|---|
| Angle of Elevation | The angle formed between the horizontal line and the observer's line of sight when looking up at an object. | tanθ=Height of Object/Distance from Observer |
| Angle of Depression | The angle formed between the horizontal line and the observer's line of sight when looking down at an object. | tanθ=Height Difference/Horizontal Distance |
Conclusion
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