Analytic Geometry & Trigonometry

Analytic Geometry

Analytic geometry expresses geometric relationships with algebra. On the F.E. exam you will use it to find distances, slopes, angles between lines, and intersections — often embedded inside physics or engineering word problems rather than as standalone “plot this curve” questions.

Distance and midpoint

Between points \(P_1(x_1, y_1)\) and \(P_2(x_2, y_2)\):

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] \[ \text{Midpoint: } \left(\frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2}\right) \]

Slope and lines

The slope of a line through two points is rise over run:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Common forms of a line:

Slope–intercept: \(y = mx + b\)
Point–slope: \(y - y_1 = m(x - x_1)\)
General: \(Ax + By + C = 0\)

Parallel lines share the same slope; perpendicular lines satisfy \(m_1 m_2 = -1\) (when neither is vertical).

Conic sections (essentials)

Know the standard forms well enough to identify a conic and extract key parameters:

Circle: \((x - h)^2 + (y - k)^2 = r^2\)
Parabola: \(y = ax^2 + bx + c\) or \((x - h)^2 = 4p(y - k)\)
Ellipse: \(\dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1\)

Trigonometry

Trigonometry relates angles to side lengths in triangles. For the F.E., right-triangle definitions and the laws of sines and cosines cover most surveying, navigation, and force-resolution problems.

Right triangles

For angle \(\theta\) in a right triangle with opposite side \(O\), adjacent side \(A\), and hypotenuse \(H\):

\[ \sin\theta = \frac{O}{H}, \quad \cos\theta = \frac{A}{H}, \quad \tan\theta = \frac{O}{A} \]

Exam habit

Sketch the triangle and label opposite, adjacent, and hypotenuse relative to the angle you are using. Most errors come from picking the wrong ratio, not from calculator mistakes.

Special angles

Memorize exact values for \(30°\), \(45°\), and \(60°\) (and their radian equivalents \(\pi/6\), \(\pi/4\), \(\pi/3\)). The reference handbook provides more, but these appear constantly.

Law of sines and law of cosines

For any triangle with sides \(a, b, c\) opposite angles \(A, B, C\):

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \quad \text{(law of sines)} \] \[ c^2 = a^2 + b^2 - 2ab\cos C \quad \text{(law of cosines)} \]

Use the law of sines when you have an angle–side pair and another angle or side. Use the law of cosines for SSS or SAS configurations.

Radians and the unit circle

Convert between degrees and radians: \(\theta_{\text{rad}} = \theta_{\text{deg}} \cdot \pi/180\). On the F.E., radian measure appears in arc length \(s = r\theta\) (with \(\theta\) in radians) and in simple harmonic motion.

\[ \sin^2\theta + \cos^2\theta = 1, \quad \tan\theta = \frac{\sin\theta}{\cos\theta} \]

Angle between lines

If two lines have slopes \(m_1\) and \(m_2\), the acute angle \(\phi\) between them satisfies:

\[ \tan\phi = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| \]

When resolving forces or velocities into components, use \(\cos\theta\) for the adjacent component and \(\sin\theta\) for the perpendicular component along a chosen axis.

Distance from a point to a line

For line \(Ax + By + C = 0\) and point \((x_0, y_0)\):

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

Example: River Width

Surveying with a transit

A surveyor stands at point C on one bank and sights point B directly across the river. Walking 250 m along the bank to point A, the angle \(\angle CAB\) is measured as 36° 30′. Find the river width \(CB\).

Right triangle \(ABC\) with the right angle at \(C\); \(CB\) is the river width.

Solution. Convert the angle to decimal degrees: \(36°30' = 36 + 30/60 = 36.5°\). The right angle is at \(C\), so \(CB\) is opposite \(\angle CAB\) and \(AC\) is adjacent:

\[ CB = AC \tan(\angle CAB) = 250 \tan(36.5°) \approx 185 \text{ m} \]

The river width is approximately 185 m (three significant figures).

Example: Force Components

Resultant along axes

A cable pulls with 500 N at 40° above the horizontal. Find the horizontal and vertical components.

Resolve \(F\) into horizontal \(F\cos\theta\) and vertical \(F\sin\theta\) components.

Solution. With the angle measured from the +x axis:

\[ F_x = 500\cos 40° \approx 383\ \text{N}, \quad F_y = 500\sin 40° \approx 321\ \text{N} \]

Check: \(\sqrt{383^2 + 321^2} \approx 500\ \text{N}\). Components are 383 N horizontal and 321 N vertical.

Work through graded problems — including a randomized river-width survey — in the dedicated practice section.

Go to Practice →