Solving Right Triangles

Core Definition

Solving a right-angled triangle means finding all unknown side lengths and angle measures. This is accomplished using the Pythagorean Theorem for side relationships:

a^2 + b^2 = c^2

and the trigonometric ratios (Sine, Cosine, and Tangent) to relate side lengths directly to acute angles.

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.

SOH CAH TOA Relationships

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

\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \qquad \cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \qquad \tan\theta = \frac{\text{Opposite}}{\text{Adjacent}}

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:

\theta = \arcsin\left(\frac{\text{Opp}}{\text{Hyp}}\right) \qquad \theta = \arccos\left(\frac{\text{Adj}}{\text{Hyp}}\right) \qquad \theta = \arctan\left(\frac{\text{Opp}}{\text{Adj}}\right)

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:

v_x = \|\vec{v}\|\cos\theta \qquad v_y = \|\vec{v}\|\sin\theta

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 $\theta$ :

x = r\cos\theta \qquad y = r\sin\theta \qquad r = \sqrt{x^2 + y^2} \qquad \theta = \arctan\left(\frac{y}{x}\right)

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 $(6, \frac{\pi}{3})$ .

Connections & References

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