In this section, we delve into the fascinating interplay between Pythagoras’ theorem and trigonometry, tools that are indispensable for solving 2D problems in mathematics. By understanding how these principles work together, students can tackle a wide range of questions with confidence, from calculating the sides of a triangle to determining the shortest distance to a line.

Pythagoras’ Theorem Recap

Pythagoras’ theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Mathematically, this is expressed as $c^2 = a^2 + b^2$, where $c$ is the length of the hypotenuse, and $a$ and $b$ are the lengths of the other two sides.

Image courtesy of Chili Math

Introduction to Trigonometry in 2D Problems

Trigonometry involves the study of the relationships between the angles and sides of triangles. The primary trigonometric ratios – sine (sin), cosine (cos), and tangent (tan) – play a pivotal role in this analysis.

• Sine of an angle in a right-angled triangle is the ratio of the length of the opposite side to the hypotenuse.
• Cosine of an angle is the ratio of the length of the adjacent side to the hypotenuse.
• Tangent of an angle is the ratio of the length of the opposite side to the adjacent side.

Combining Pythagoras and Trigonometry

Combining these principles allows for solving complex problems involving right-angled triangles. Let's explore how these can be applied to calculate unknown sides and angles, and understand the concept of the shortest distance to a line through examples.

Example 1: Calculating Sides in a Right-Angled Triangle

Problem: In a right-angled triangle, one angle is $30^\circ$, and the hypotenuse is 10 cm long. Calculate the lengths of the opposite and adjacent sides.

Solution:

1. Calculate the Opposite Side using Sine:

$\sin(30^\circ) = \dfrac{\text{Opposite}}{\text{Hypotenuse}}$

$\dfrac{1}{2} = \dfrac{\text{Opposite}}{10}$

$\text{Opposite} = 5 \text{ cm}$

2. Calculate the Adjacent Side using Pythagoras’ Theorem:

$c^2 = a^2 + b^2$

$10^2 = 5^2 + b^2$

$100 = 25 + b^2$

$b^2 = 75$

$b = \sqrt{75}$

$b = \approx 8.66 \text{ cm}$

Example 2: Finding the Shortest Distance to a Line

Problem: A point is located 8 cm away from a line at an angle of elevation of $45^\circ$. Calculate the shortest distance from the point to the line.

Solution:

1. Understand the Scenario: The shortest distance from a point to a line in this context is the length of the perpendicular (opposite side) dropped from the point to the line.

2. Use Trigonometry:

$\tan(45^\circ) = \dfrac{\text{Opposite}}{\text{Adjacent}}$

Since $\tan(45^\circ) = 1$,

$1 = \frac{\text{Opposite}}{8}$

$\text{Opposite} = 8 \text{ cm}$

The shortest distance to the line is 8 cm, demonstrating how trigonometry can directly provide solutions without the explicit need for Pythagoras’ theorem in this case.

Practice Questions

To solidify your understanding, try solving these practice questions:

Question 1:

A right-angled triangle has one angle of $60^\circ$ and an adjacent side of 5 cm. Calculate the length of the hypotenuse and the opposite side.

Solution:

1. Hypotenuse $(c)$:

$c = \dfrac{5}{\cos(60^\circ)}$

Since $\cos(60^\circ) = \dfrac{1}{2}$,

we have: $c = 5 \times 2 = 10 \text{ cm}$

2. Opposite side $(b)$:

$b = 5 \times \tan(60^\circ)$

Given that $\tan(60^\circ) = \sqrt{3}$,

it simplifies to: $b = 5 \times \sqrt{3}$

Therefore, the hypotenuse is $10 \text{ cm}$ and the opposite side is $5\sqrt{3} \text{ cm}$.

Question 2:

Determine the shortest distance from a point to a line, given the point is 6 cm from the line at an angle of depression of $30^\circ$.

Solution:

1. Given Data:

• Angle of depression: $30^\circ$
• Distance from the point to the line at this angle (hypotenuse): 6 cm

2. Objective:

• Find the shortest distance from the point to the line, which corresponds to the perpendicular (vertical) distance in the right-angled triangle formed.

3. Use of Trigonometry:

• In a right-angled triangle, the sine of an angle $(\sin)$ is the ratio of the length of the side opposite to the angle (which is the shortest distance in our case) to the length of the hypotenuse (the given distance of 6 cm).
• The sine of $30^\circ$ is a well-known value: $\sin(30^\circ) = \dfrac{1}{2}$.

4. Calculation:

To find the shortest distance, we multiply the hypotenuse (6 cm) by $\sin(30^\circ)$:

$\text{Shortest distance} = 6 \text{ cm} \times \sin(30^\circ)$

$\text{Shortest distance} = 6 \text{ cm} \times \frac{1}{2}$

$\text{Shortest distance} = 3 \text{ cm}$

Thus, the shortest distance from the point to the line, based on the given angle of depression and distance, is calculated to be 3 cm.

Tips for Solving Problems

• Identify Known Quantities: Always start by identifying which sides and angles you know.
• Select the Right Formula: Choose whether Pythagoras’ theorem or trigonometric ratios (or both) are needed based on what you’re solving for.
• Solve Step by Step: Break down the problem into smaller steps and solve each part carefully.