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# Benchmark Angles|Definition & Meaning

## Definition

**Benchmark angles** are such angles that are used as a **reference** or **standard** to estimate other angles such as **acute** and **obtuse** angles closer to them. These angles have exact values of **trigonometric ratios** which makes the calculations easier with chances of fewer errors.

**Figure 1** shows a demonstration of **benchmark** angles.

## Introduction to the Terminology

The word “**Benchmark**” refers to a known **standard** that sets its values to be analyzed, compared, and measured against other values that are not **commonly** used.

An **angle** consists of** two rays** whose endpoints are joined at a point known as the **vertex**.

## Types of Benchmark Angles

Benchmark angles play an essential role in **trigonometry** and can be divided into **four** major types. These are the **quadrantal** angles, the **30°**, **45°**, and **60°** angles.

All the other benchmark angles are the **reference angles** of **30°**, **45°**, and **60° **angles. These types of Benchmark angles are explained below.

### Quadrantal Angles

**Quadrantal angles** are the angles whose one ray lies on the **coordinate axes** which are the **x** and **y** axes. They lie on the **boundaries** of the four quadrants.

These are **0°**, **90°**, **180°**, and **270°** with one ray lying on +x-axis, +y-axis, -x-axis, and -y-axis respectively. In radians, the quadrantal angles are **0**, **π/2**, **π**, and **3π/2** rad.

The angles such as **360° **co-terminal to **0° **and other **co-terminal** angles are also considered **quadrantal** angles.

**Figure 2** shows the **four** quadrantal angles.

The (**cosine**, **sine**) values for the quadrantal angles** 0°**, **90°**, **180°**, and **270°** are (**1**,**0**), (**0**,**1**), (**-1**,**0**), and (**0**,**-1**) respectively.

### 30**°** Angle and Its Reference Angles

The **30°** angle is a **benchmark** angle and is commonly used as a reference. It is equal to **π/3** radians. It is an acute angle and lies in the **first** quadrant.

The reference angles of **30°** are **150°**, **210°**, and **330°** in the second, third and fourth quadrants.

In **radians**, these angles are equal to **5π/6**, **7π/6**, and **11π/6** rad respectively. These are obtained from the half (**180°**) or full (**360°**) rotation along a circle.

They can be written as follows:

**150° = 180° – 30°**

**210° = 180° + 30°**

**330° = 360° – 30°**

Benchmark angle **of 30°** and its **reference** angles are shown in **figure 3**.

### 45**°** Angle and Its Reference Angles

The **45°** angle is acute and is also a **benchmark** angle. It is equal to **π/4** radians and lies in the first quadrant.

The **45°** angle has reference angles of **135°**, **225°**, and **315°** in the second, third and fourth quadrants.

These are obtained by subtracting **45°** from **180°**, adding **45°** to **180°**, and subtracting **45°** from **360°**.

These **reference angles** can be written with the **45°** angles as follows:

**135° = 180° – 45°**

**225° = 180° + 45°**

**315° = 360° – 45°**

The reference angles of **45°** in radians are **3π/4**, **5π/4**, and **7π/4** **rad**. They are all the multiples of **π/4** which is **45°** in radians. These are shown in **figure 4**.

### 60**°** Angle and Its Reference Angles

The **60°** angle is a benchmark angle and is commonly used in **trigonometric** calculations. It is equal to **π/3** radians found in the first quadrant.

The reference angles of **60°** degrees are **120°**, **240°**, and **300° **degrees. They are also **benchmark** angles as they are used as a standard in different trigonometric problems.

They are equal to **2π/3**, **4π/3**, and **5π/3** radians. All have a common multiple of** π/3** radians. The reference angles can be written with **60°** degrees added or subtracted as follows:

**120° = 180° – 60°**

**240° = 180° + 60°**

**300° = 360° – 60°**

**Figure 5** shows the angle **of 60°** along with its **reference** angles.

## Use of Benchmark Angles

The** benchmark angles** are used in the approximation of **acute** angles and **obtuse** angles. This is done through the **quadrantal** benchmark angles which are **0°** and **90°**.

### Approximating Acute Angles

**Acute** angles are such angles that lie between **0°** and** 90°**. The **benchmark** angles **30°**, **45°,** and **60°** are also acute angles. Any acute angle other than the benchmark angles can be **approximated** using the two quadrant angles **0°** and **90°**.

At first, the unknown angle is **compared** with the two **quadrantal** angles in such a way as to find which quadrantal angle is nearest to the **unknown angle**.

If the angle is near **0°**, the angle can be estimated to be **15°**. This is reasonable as it is near to **0°** than to **90°**. **Figure** **6** shows the approximated **15°** angle lying near **0°**.

Similarly, if the **acute** angle is closer to** 90°**, the angle can be estimated to be **85°** as shown in **figure 7**.

### Approximating Obtuse Angles

**Obtuse** angles are the angles present between **90°** and **180°**. The obtuse angles can be estimated by checking the closeness of the given angle to the two **quadrantal** angles, **90°** and **180°**.

For **example**, if the angle is closer to **90°** and is **obtuse**, the estimated angle will be **100°**. This is shown in **figure 8**.

These **approximations** help determine how **large** or **small** an angle is from the **benchmark** angles.

## Using Benchmark Angles in a Problem

What is **cos(A + B)** if A and B are the **benchmark** angles where **A** = **30°** and **B** = **60°**? Also, prove the trigonometric property:

**cos(A + B) = cos(A).cos(B) – sin(A).sin(B)**

### Solution

At first, putting the benchmark angles in **cos(A + B)** gives the result as follows:

cos(A + B) = cos(30° + 60°) = cos(90°) = 0

Now, putting the values on the** right-hand side** of the equation gives:

R.H.S = cos(A).cos(B) – sin(A).sin(B)

R.H.S = cos(30°).cos(60°) – sin(30°).sin(60°)

\[ R.H.S = \Big\{ \frac{ \sqrt{3} }{2} \Big\} . \Big\{ \frac{1}{2} \Big\} \ – \ \Big\{ \frac{1}{2} \Big\} .\Big\{ \frac{ \sqrt{3} }{2}\Big\} \]

\[ R.H.S = \Big\{ \frac{ \sqrt{3} }{4} \Big\} \ – \ \Big\{ \frac{ \sqrt{3} }{4} \Big\} \]

R.H.S = 0

As we can see that:

L.H.S = cos(A + B) = 0

L.H.S = R.H.S

Hence **proved**.

*All the images are created using GeoGebra.*