Solving Right Triangles

Concept Explanation

In any right triangle, one angle is always $90^\circ$ (the right angle), and the side opposite this right angle is the hypotenuse ($c$, the longest side). The other two sides are labeled opposite and adjacent relative to the specific acute angle $\theta$ you are analyzing.

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SOH CAH TOA Relationships

The three primary trigonometric ratios can be remembered using the acronym SOH CAH TOA:

1. Finding an Unknown Side (Given a Side and an Angle)

If you know one acute angle and one side length, you can isolate the unknown side using algebraic multiplication:

• Finding Opposite: $\text{Opposite} = \text{Hypotenuse} \cdot \sin\theta$
• Finding Adjacent: $\text{Adjacent} = \text{Hypotenuse} \cdot \cos\theta$

2. Finding an Unknown Angle (Given Two Sides)

If you know two side lengths, you can solve for the acute angle $\theta$ using the inverse trigonometric functions:

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Why We Use It in Multivariable Calculus (Calculus III)

Solving right triangles is the absolute foundation of multidimensional math:

1. Vector Decomposition

Any 2D vector $\vec{v}$ can be viewed as the hypotenuse of a right triangle whose legs are the horizontal component $v_x$ and vertical component $v_y$. Resolving a vector into its components means "solving" this right triangle:

In 3D, we use right triangles to decompose vectors in spherical coordinates using the polar angle $\theta$ and the azimuthal angle $\phi$.

2. Coordinate Transformations

Converting coordinates between Cartesian $(x,y)$ and Polar $(r,\theta)$ (or Cylindrical and Spherical systems in 3D) is mathematically identical to solving a right triangle with legs $x$ and $y$, hypotenuse $r$, and angle $\theta$:

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Example Problems

Example Problem 1: Vector Decomposition (Finding Sides)

A wind velocity vector $\vec{w}$ has a magnitude of $12\text{ m/s}$ and blows at an angle of $30^\circ$ north of east (equivalent to an angle of $30^\circ$ relative to the positive $x$-axis). Decompose this velocity vector into its eastern component $w_x$ and northern component $w_y$.

Solution

Step 1: Set up the right triangle The vector $\vec{w}$ forms the hypotenuse of a right triangle:

• Hypotenuse $\|\vec{w}\| = 12$
• Angle $\theta = 30^\circ$
• Eastern component $w_x$ is the adjacent side.
• Northern component $w_y$ is the opposite side.

Step 2: Solve for the adjacent side ($w_x$)

$$w_x = \|\vec{w}\|\cos(30^\circ) = 12 \cdot \frac{\sqrt{3}}{2} = 6\sqrt{3} \approx 10.39\text{ m/s}$$

Step 3: Solve for the opposite side ($w_y$)

$$w_y = \|\vec{w}\|\sin(30^\circ) = 12 \cdot \frac{1}{2} = 6\text{ m/s}$$

The component form of the vector is $\vec{w} = \langle 6\sqrt{3}, 6 \rangle$.

Example Problem 2: Coordinate Conversion (Finding Angles & Hypotenuse)

A robotic arm is tracking a target at the Cartesian coordinates $(3, 3\sqrt{3})$. Find the polar coordinates $(r, \theta)$ of the target to orient the arm.

Solution

Step 1: Identify legs of the right triangle

• Adjacent leg $x = 3$
• Opposite leg $y = 3\sqrt{3}$

Step 2: Find the hypotenuse ($r$) using the Pythagorean Theorem

$$r = \sqrt{x^2 + y^2} = \sqrt{3^2 + (3\sqrt{3})^2} = \sqrt{9 + (9 \cdot 3)} = \sqrt{9 + 27} = \sqrt{36} = 6$$

Step 3: Find the angle ($\theta$) using the inverse tangent

$$\theta = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{3\sqrt{3}}{3}\right) = \arctan(\sqrt{3})$$

Since $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$ (or $60^\circ$):

$$\theta = \frac{\pi}{3} \text{ radians} \quad (60^\circ)$$

The target's polar coordinates are $(6, \frac{\pi}{3})$.

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Connections & References

• Parent Concepts: Calculus Foundations
• Related Concepts: Parametric Functions, Dot and Cross Products
• Course Links: Calc-3 Discussion 01