Plane Trigonometry

LessonAbout 1.5 h1 of 2 in Plane and Spherical Trigonometry

Trigonometric Functions of Right Triangles

In a right triangle, the six trigonometric functions relate the acute interior angles to the ratios of the lengths of the opposite side, adjacent side, and hypotenuse. The complementary angle relationships state that any trigonometric function of an acute angle equals the co-function of its complement.

Fundamental right triangle function ratios:
\[ \sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}, \qquad\qquad \cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}, \qquad\qquad \tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} \]
\[ \csc\theta = \dfrac{\text{hypotenuse}}{\text{opposite}}, \qquad\qquad \sec\theta = \dfrac{\text{hypotenuse}}{\text{adjacent}}, \qquad\qquad \cot\theta = \dfrac{\text{adjacent}}{\text{opposite}} \]
Pythagorean Theorem:
\[ a^2 + b^2 = c^2 \]
Co-function identities for complementary angles:
\[ \sin\theta = \cos(90^\circ - \theta), \qquad\qquad \tan\theta = \cot(90^\circ - \theta), \qquad\qquad \sec\theta = \csc(90^\circ - \theta) \]
Coversed sine (coversine):
\[ \text{coversin}\theta = 1 - \sin\theta \]
  • Six basic trigonometric ratios define right triangle sides
  • Pythagorean theorem relates legs to the hypotenuse
  • Co-functions of complementary angles are strictly equal
  • Coversed sine equals one minus sine of angle

Worked Example 1: Evaluating functions given one ratio

Given an acute angle \(A\) with \(\sin A = \dfrac{3}{7}\), find the exact values of \(\cos A\) and \(\tan A\).

Solution: In a right triangle with opposite leg \(a = 3\) and hypotenuse \(c = 7\), find adjacent leg \(b\) using the Pythagorean theorem:

\[ b = \sqrt{c^2 - a^2} = \sqrt{7^2 - 3^2} = \sqrt{40} = 2\sqrt{10} \]
\[ \cos A = \dfrac{b}{c} = \dfrac{2\sqrt{10}}{7} \]
\[ \tan A = \dfrac{a}{b} = \dfrac{3}{2\sqrt{10}} = \dfrac{3\sqrt{10}}{20} \]

Worked Example 2: Solving co-function equations

If \(\sin 3A = \cos 6B\), determine the linear equation relating angles \(A\) and \(B\).

Solution: Express cosine in terms of its sine co-function: \(\cos 6B = \sin(90^\circ - 6B)\).

\[ \sin 3A = \sin(90^\circ - 6B) \implies 3A = 90^\circ - 6B \]
\[ 3A + 6B = 90^\circ \implies A + 2B = 30^\circ \]

Worked Example 3: Finding an angle from coversed sine

If \(\text{coversin}\theta = 0.134\), solve for angle \(\theta\).

Solution: Apply the definition \(\text{coversin}\theta = 1 - \sin\theta\):

\[ 1 - \sin\theta = 0.134 \implies \sin\theta = 0.866 \]
\[ \theta = \arcsin(0.866) = 60^\circ \]

Trigonometric Identities and Equations

Trigonometric identities express relationships that hold true for all permissible angle values. Algebraic equations involving trigonometric functions can be simplified or solved using fundamental identities.

Pythagorean Relations:
\[ \sin^2 A + \cos^2 A = 1 \]
\[ 1 + \tan^2 A = \sec^2 A \]
\[ 1 + \cot^2 A = \csc^2 A \]
Sum and Difference Formulas:
\[ \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B \]
\[ \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B \]
\[ \tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B} \]
Double-Angle Formulas:
\[ \sin 2A = 2\sin A \cos A \]
\[ \cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A \]
\[ \tan 2A = \dfrac{2\tan A}{1 - \tan^2 A} \]
Half-Angle Formulas:
\[ \sin\dfrac{A}{2} = \pm\sqrt{\dfrac{1 - \cos A}{2}}, \qquad\qquad \cos\dfrac{A}{2} = \pm\sqrt{\dfrac{1 + \cos A}{2}} \]
  • Pythagorean identities connect square of fundamental functions
  • Sum and difference identities expand combined angle functions
  • Double angle formulas express functions of doubled argument
  • Half angle formulas compute functions of half angle

Worked Example 4: Simplifying trigonometric expressions

Simplify the expression \(\dfrac{\sin\theta + \cos\theta\tan\theta}{\cos\theta}\).

Solution: Substitute \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\) into the numerator:

\[ \dfrac{\sin\theta + \cos\theta\left(\dfrac{\sin\theta}{\cos\theta}\right)}{\cos\theta} = \dfrac{\sin\theta + \sin\theta}{\cos\theta} = \dfrac{2\sin\theta}{\cos\theta} = 2\tan\theta \]

Worked Example 5: Solving a trigonometric quadratic equation

Find the value of angle \(A\) between \(270^\circ\) and \(360^\circ\) satisfying \(2\sin^2 A - \sin A = 1\).

Solution: Rearrange into standard quadratic form \(2\sin^2 A - \sin A - 1 = 0\) and factor:

\[ (2\sin A + 1)(\sin A - 1) = 0 \]

Setting factors to zero gives \(\sin A = 1\) (which gives \(A = 90^\circ\)) or \(\sin A = -0.5\). In the fourth quadrant (\(270^\circ < A < 360^\circ\)):

\[ A = 360^\circ - 30^\circ = 330^\circ \]

Worked Example 6: Inverse trigonometric equation

Solve for \(x\) in the equation \(\arctan(2x) + \arctan x = \dfrac{\pi}{4}\).

Solution: Take the tangent of both sides using the sum formula \(\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}\):

\[ \dfrac{2x + x}{1 - (2x)(x)} = \tan\dfrac{\pi}{4} \implies \dfrac{3x}{1 - 2x^2} = 1 \]
\[ 3x = 1 - 2x^2 \implies 2x^2 + 3x - 1 = 0 \]

Apply the quadratic formula:

\[ x = \dfrac{-3 + \sqrt{3^2 - 4(2)(-1)}}{2(2)} = \dfrac{-3 + \sqrt{17}}{4} = \dfrac{-3 + 4.123}{4} = 0.281 \]

Oblique Triangles and Solutions

An oblique triangle has no right angle. It can be solved using the Law of Sines when given two angles and a side or two sides and an opposite angle, or the Law of Cosines when given three sides or two sides and their included angle.

Law of Sines:
\[ \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R \]
Law of Cosines:
\[ a^2 = b^2 + c^2 - 2bc\cos A \]
\[ b^2 = a^2 + c^2 - 2ac\cos B \]
\[ c^2 = a^2 + b^2 - 2ab\cos C \]
Law of Tangents:
\[ \dfrac{a - b}{a + b} = \dfrac{\tan\frac{1}{2}(A - B)}{\tan\frac{1}{2}(A + B)} \]
Median length formula to side \(b\):
\[ m_b = \dfrac{1}{2}\sqrt{2a^2 + 2c^2 - b^2} \]
  • Law of Sines relates sides to opposite angle sines
  • Law of Cosines solves triangles given three sides
  • Law of Cosines solves triangles given two sides included angle
  • Medians bisect opposite sides from corresponding vertices

Worked Example 7: Solving side using Law of Cosines (SAS)

In triangle \(ABC\), side \(a = 15\text{ m}\), side \(b = 20\text{ m}\), and angle \(C = 100^\circ\). Solve for side \(c\).

Solution: Apply the Law of Cosines for side \(c\):

\[ c^2 = a^2 + b^2 - 2ab\cos C = (15)^2 + (20)^2 - 2(15)(20)\cos 100^\circ \]
\[ c^2 = 225 + 400 - 600(-0.17365) = 625 + 104.19 = 729.19 \]
\[ c = \sqrt{729.19} = 27.01\text{ m} \]

Worked Example 8: Median length of an oblique triangle

In triangle \(ABC\), \(a = 18\text{ cm}\), \(b = 24\text{ cm}\), and \(c = 34\text{ cm}\). Find the length of the median drawn from vertex \(B\) to side \(b\).

Solution: Apply the median formula \(m_b = \dfrac{1}{2}\sqrt{2a^2 + 2c^2 - b^2}\):

\[ m_b = \dfrac{1}{2}\sqrt{2(18)^2 + 2(34)^2 - (24)^2} = \dfrac{1}{2}\sqrt{2(324) + 2(1156) - 576} \]
\[ m_b = \dfrac{1}{2}\sqrt{648 + 2312 - 576} = \dfrac{1}{2}\sqrt{2384} = \dfrac{48.826}{2} = 24.41\text{ cm} \]

Area of Plane Triangles and Quadrilaterals

The area of a triangle can be calculated using base and altitude, two sides and their included angle, or three sides using Heron’s formula. Area relationships also determine the radii of inscribed and circumscribed circles.

Area given base \(b\) and altitude \(h\):
\[ A = \dfrac{1}{2}bh \]
Area given two sides and included angle:
\[ A = \dfrac{1}{2}ab\sin\theta \]
Heron’s Formula (given three sides):
\[ A = \sqrt{s(s-a)(s-b)(s-c)}, \qquad\qquad s = \dfrac{a+b+c}{2} \]
Inscribed circle radius (\(r\)):
\[ A = rs \implies r = \dfrac{A}{s} \]
Circumscribed circle radius (\(R\)):
\[ A = \dfrac{abc}{4R} \implies R = \dfrac{abc}{4A} \]
Area of quadrilateral given diagonals \(d_1, d_2\) and angle \(\theta\):
\[ A = \dfrac{1}{2}d_1 d_2 \sin\theta \]
Cyclic quadrilateral area (Brahmagupta’s Formula):
\[ A = \sqrt{(s-a)(s-b)(s-c)(s-d)}, \qquad\qquad s = \dfrac{a+b+c+d}{2} \]
Ptolemy’s Theorem for cyclic quadrilaterals:
\[ ac + bd = d_1 d_2 \]
  • Heron’s formula determines triangle area from side lengths alone
  • Inradius equals triangle area divided by semiperimeter
  • Circumradius relates product of sides to four times area
  • Brahmagupta’s formula calculates cyclic quadrilateral area

Worked Example 9: Area and inscribed circle radius

A triangle has sides \(a = 13\text{ cm}\), \(b = 14\text{ cm}\), and \(c = 15\text{ cm}\). Calculate its area and the radius of its inscribed circle.

Solution: Calculate semi-perimeter \(s = \dfrac{13 + 14 + 15}{2} = 21\text{ cm}\). Apply Heron’s formula:

\[ A = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21(8)(7)(6)} = \sqrt{7056} = 84\text{ cm}^2 \]

Find inradius \(r\) using \(A = rs\):

\[ r = \dfrac{A}{s} = \dfrac{84}{21} = 4\text{ cm} \]

Worked Example 10: Circumscribed circle radius

Find the radius \(R\) of the circle circumscribing the triangle with sides \(a = 8\text{ cm}\), \(b = 10\text{ cm}\), and \(c = 14\text{ cm}\), given its area is \(39.19\text{ cm}^2\).

Solution: Use the circumscribed radius formula \(A = \dfrac{abc}{4R}\):

\[ 39.19 = \dfrac{(8)(10)(14)}{4R} = \dfrac{1120}{4R} = \dfrac{280}{R} \]
\[ R = \dfrac{280}{39.19} = 7.14\text{ cm} \]

Angles of Elevation, Depression, and Bearings

Angles of elevation and depression are acute angles formed by a horizontal line of sight and the observer’s line of vision to an object above or below the horizontal line, respectively.

Height and distance relationship:
\[ \tan\theta = \dfrac{\text{vertical height } h}{\text{horizontal distance } d} \]
Height formula for two observations along a straight horizontal line:
\[ h = \dfrac{d}{\cot\alpha - \cot\beta} = \dfrac{d\tan\alpha\tan\beta}{\tan\beta - \tan\alpha} \]
Speed and distance relation:
\[ \text{Distance } d = \text{Speed } v \times \text{Time } t \]
  • Angles of elevation are measured upward from horizontal plane
  • Angles of depression are measured downward from horizontal plane
  • Two-point sightings yield tower height without measuring base distance
  • Bearings indicate direction measured clockwise from North or South

Worked Example 11: Tower height from two angle measurements

An observer finds the angle of elevation of the top of a tower to be \(30^\circ\). After walking \(85\text{ m}\) closer to the tower along level ground, the angle of elevation becomes \(60^\circ\). Find the height \(h\) of the tower.

Solution: Let \(x\) be the distance from the second observation point to the base of the tower.

\[ \tan 60^\circ = \dfrac{h}{x} \implies h = x\tan 60^\circ = x\sqrt{3} \]
\[ \tan 30^\circ = \dfrac{h}{85 + x} \implies h = (85 + x)\tan 30^\circ = \dfrac{85 + x}{\sqrt{3}} \]

Equating the two expressions for \(h\):

\[ x\sqrt{3} = \dfrac{85 + x}{\sqrt{3}} \implies 3x = 85 + x \implies 2x = 85 \implies x = 42.5\text{ m} \]
\[ h = 42.5\tan 60^\circ = 42.5(1.732) = 73.61\text{ m} \]

Worked Example 12: Leaning pole shadow problem

A pole leans \(15^\circ\) from the vertical directly toward the sun. When the angle of elevation of the sun is \(61^\circ\), the pole casts a shadow of \(15\text{ m}\) on level ground. Determine the length of the pole \(x\).

Solution: Construct the triangle formed by ground shadow, pole, and sun ray. Because the pole leans toward the sun, the shadow lies on the opposite side of the pole’s base. The interior angles of the triangle are:

Angle at ground shadow tip \(= 61^\circ\).

Angle at base of pole \(= 90^\circ + 15^\circ = 105^\circ\).

Angle at top of pole \(= 180^\circ - (61^\circ + 105^\circ) = 14^\circ\).

Apply Law of Sines to solve for pole length \(x\):

\[ \dfrac{x}{\sin 61^\circ} = \dfrac{15}{\sin 14^\circ} \implies x = \dfrac{15\sin 61^\circ}{\sin 14^\circ} = \dfrac{15(0.8746)}{0.2419} = 54.2\text{ m} \]

Formula Summary: Plane Trigonometry

Formula Name Formula
Sine Ratio \(\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}\)
Cosine Ratio \(\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}\)
Tangent Ratio \(\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}\)
Pythagorean Theorem \(a^2 + b^2 = c^2\)
Complementary Angle Co-function \(\sin\theta = \cos(90^\circ - \theta)\)
Coversed Sine \(\text{coversin}\theta = 1 - \sin\theta\)
Pythagorean Identity \(\sin^2 A + \cos^2 A = 1\)
Secant-Tangent Identity \(1 + \tan^2 A = \sec^2 A\)
Sine Sum and Difference \(\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B\)
Cosine Sum and Difference \(\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B\)
Double-Angle Sine \(\sin 2A = 2\sin A \cos A\)
Double-Angle Cosine \(\cos 2A = \cos^2 A - \sin^2 A = 1 - 2\sin^2 A\)
Law of Sines \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R\)
Law of Cosines \(c^2 = a^2 + b^2 - 2ab\cos C\)
Median Length \(m_b = \dfrac{1}{2}\sqrt{2a^2 + 2c^2 - b^2}\)
Heron’s Formula \(A = \sqrt{s(s-a)(s-b)(s-c)}\)
Inscribed Circle Radius \(r = \dfrac{A}{s}\)
Circumscribed Circle Radius \(R = \dfrac{abc}{4A}\)
Triangle Area (Two Sides and Angle) \(A = \dfrac{1}{2}ab\sin\theta\)
Quadrilateral Diagonal Area \(A = \dfrac{1}{2}d_1 d_2 \sin\theta\)
Brahmagupta’s Formula \(A = \sqrt{(s-a)(s-b)(s-c)(s-d)}\)
Two-Observation Height Formula \(h = \dfrac{d}{\cot\alpha - \cot\beta}\)