Engineering Mathematics II for Mechatronics Technology professionals in Kenya requires a strong grasp of vectors and their applications, especially in analyzing forces, motions, and fields in mechatronic systems. This chapter focuses on applying vector theorems to angles, including acute, obtuse, reflex, and right angles, which are fundamental in resolving vector components, calculating resultant forces, and designing systems with precise angular relationships. Mastery of these concepts enables technicians and engineers to model and solve real-world problems such as robotic arm positioning, sensor alignment, and motion control in automated systems.

2.1 Angles and hyperbolic

Angles are a core aspect of vector analysis in mechatronics, where components often interact at various angular orientations. Understanding the types of angles, acute, obtuse, reflex, and right, is essential for correctly applying vector theorems to resolve forces and motions. This section covers the definitions, formulas, and calculations related to these angles, providing a foundation for advanced vector operations in mechatronic applications.

2.1.1 Acute functions

Acute angles are those less than 90°, frequently encountered when analyzing joint angles in robotic arms or sensor fields of view. The trigonometric functions for acute angles, sine, cosine, and tangent, are fundamental for decomposing vectors into components and calculating magnitudes in mechatronic devices.

The basic trigonometric functions for an acute angle \(\theta\) are:

$$ \sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}}, \quad \cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}}, \quad \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} $$

Worked Examples

Example 1: A robotic arm joint forms an angle of 30° with the horizontal. If the arm length is 1.5 m, find the vertical and horizontal components of the arm's reach.

Given: \(\theta = 30^\circ\), hypotenuse \(= 1.5\) m

$$ \text{Vertical component} = 1.5 \times \sin 30^\circ $$$$ = 1.5 \times 0.5 $$$$ = 0.75 \text{ m} $$

$$ \text{Horizontal component} = 1.5 \times \cos 30^\circ $$$$ = 1.5 \times 0.866 $$$$ = 1.299 \text{ m} $$

Answer: Vertical = 0.75 m, Horizontal = 1.299 m

Example 2: A sensor detects an object at an angle of 45° to its baseline. If the sensor's detection range is 3 m, find the components of the detection vector along the baseline and perpendicular to it.

Given: \(\theta = 45^\circ\), hypotenuse \(= 3\) m

$$ \text{Baseline component} = 3 \times \cos 45^\circ $$$$ = 3 \times 0.707 $$$$ = 2.121 \text{ m} $$

$$ \text{Perpendicular component} = 3 \times \sin 45^\circ $$$$ = 3 \times 0.707 $$$$ = 2.121 \text{ m} $$

Answer: Baseline = 2.121 m, Perpendicular = 2.121 m

Example 3: A conveyor belt moves at an angle of 60° to the horizontal with speed 4 m/s. Find the horizontal and vertical speed components.

Given: \(\theta = 60^\circ\), speed \(= 4\) m/s

$$ v_x = 4 \times \cos 60^\circ $$$$ = 4 \times 0.5 $$$$ = 2 \text{ m/s} $$

$$ v_y = 4 \times \sin 60^\circ $$$$ = 4 \times 0.866 $$$$ = 3.464 \text{ m/s} $$

Answer: Horizontal speed = 2 m/s, Vertical speed = 3.464 m/s

2.1.2 Obtuse

Obtuse angles range between 90° and 180° and appear in mechatronic linkages when components extend beyond right angles. Calculating vector components with obtuse angles requires careful application of trigonometric functions considering the angle's supplement.

For an obtuse angle \(\theta\), use the identity:

$$ \sin \theta = \sin (180^\circ - \theta), \quad \cos \theta = -\cos (180^\circ - \theta) $$

This ensures correct sign assignment for components in vector analysis.

Worked Examples

Example 1: A robotic arm segment forms an angle of 120° with the base. The arm is 2 m long. Find the horizontal and vertical components relative to the base.

Given: \(\theta = 120^\circ\), hypotenuse \(= 2\) m

Calculate reference angle:

$$ 180^\circ - 120^\circ = 60^\circ $$

$$ \text{Horizontal component} = 2 \times \cos 120^\circ = 2 \times (-\cos 60^\circ) $$$$ = 2 \times (-0.5) $$$$ = -1.0 \text{ m} $$

$$ \text{Vertical component} = 2 \times \sin 120^\circ = 2 \times \sin 60^\circ $$$$ = 2 \times 0.866 $$$$ = 1.732 \text{ m} $$

Answer: Horizontal = -1.0 m, Vertical = 1.732 m

Example 2: A force of 150 N acts at an angle of 135° from the positive x-axis. Find the x and y components of the force.

Given: \(F = 150\) N, \(\theta = 135^\circ\)

Reference angle:

$$ 180^\circ - 135^\circ = 45^\circ $$

$$ F_x = 150 \times \cos 135^\circ = 150 \times (-\cos 45^\circ) $$$$ = 150 \times (-0.707) $$$$ = -106.05 \text{ N} $$

$$ F_y = 150 \times \sin 135^\circ = 150 \times \sin 45^\circ $$$$ = 150 \times 0.707 $$$$ = 106.05 \text{ N} $$

Answer: \(F_x = -106.05\) N, \(F_y = 106.05\) N

Example 3: An actuator applies a thrust of 500 N at 150° to the horizontal axis. Calculate the components of the thrust.

Given: \(F = 500\) N, \(\theta = 150^\circ\)

Reference angle:

$$ 180^\circ - 150^\circ = 30^\circ $$

$$ F_x = 500 \times \cos 150^\circ = 500 \times (-\cos 30^\circ) $$$$ = 500 \times (-0.866) $$$$ = -433 \text{ N} $$

$$ F_y = 500 \times \sin 150^\circ = 500 \times \sin 30^\circ $$$$ = 500 \times 0.5 $$$$ = 250 \text{ N} $$

Answer: \(F_x = -433\) N, \(F_y = 250\) N

2.1.3 Reflex

Reflex angles exceed 180° but are less than 360°. In mechatronics, reflex angles occur in complex joint rotations or sensor sweep ranges. Vector calculations involving reflex angles use the difference from 360° for reference.

For reflex angle \(\theta\):

$$ \sin \theta = -\sin (360^\circ - \theta), \quad \cos \theta = \cos (360^\circ - \theta) $$

This guides correct sign assignment in vector component calculations.

Worked Examples

Example 1: A rotating arm extends at 210° from the reference axis, length 1.2 m. Find the x and y components.

Given: \(\theta = 210^\circ\), length \(= 1.2\) m

Reference angle:

$$ 360^\circ - 210^\circ = 150^\circ $$

$$ x = 1.2 \times \cos 210^\circ = 1.2 \times \cos 150^\circ $$$$ = 1.2 \times (-0.866) $$$$ = -1.039 \text{ m} $$

$$ y = 1.2 \times \sin 210^\circ = 1.2 \times (-\sin 150^\circ) $$$$ = 1.2 \times (-0.5) $$$$ = -0.6 \text{ m} $$

Answer: \(x = -1.039\) m, \(y = -0.6\) m

Example 2: A sensor beam points at 270°. The beam length is 4 m. Determine the components.

Given: \(\theta = 270^\circ\), length \(= 4\) m

Reference angle:

$$ 360^\circ - 270^\circ = 90^\circ $$

$$ x = 4 \times \cos 270^\circ = 4 \times \cos 90^\circ $$$$ = 4 \times 0 $$$$ = 0 \text{ m} $$

$$ y = 4 \times \sin 270^\circ = 4 \times (-\sin 90^\circ) $$$$ = 4 \times (-1) $$$$ = -4 \text{ m} $$

Answer: \(x = 0\) m, \(y = -4\) m

Example 3: A force of 250 N acts at 300°. Find the components.

Given: \(F = 250\) N, \(\theta = 300^\circ\)

Reference angle:

$$ 360^\circ - 300^\circ = 60^\circ $$

$$ F_x = 250 \times \cos 300^\circ = 250 \times \cos 60^\circ $$$$ = 250 \times 0.5 $$$$ = 125 \text{ N} $$

$$ F_y = 250 \times \sin 300^\circ = 250 \times (-\sin 60^\circ) $$$$ = 250 \times (-0.866) $$$$ = -216.5 \text{ N} $$

Answer: \(F_x = 125\) N, \(F_y = -216.5\) N

2.1.4 Right angle

Right angles (90°) are fundamental in mechatronics for orthogonal component analysis, sensor calibration, and coordinate transformations. The trigonometric functions at 90° simplify vector calculations.

Key values:

$$ \sin 90^\circ = 1, \quad \cos 90^\circ = 0, \quad \tan 90^\circ \text{ is undefined} $$

Worked Examples

Example 1: A force of 100 N acts vertically upward (90°). Find the horizontal and vertical components.

Given: \(F = 100\) N, \(\theta = 90^\circ\)

$$ F_x = 100 \times \cos 90^\circ = 100 \times 0 = 0 \text{ N} $$

$$ F_y = 100 \times \sin 90^\circ = 100 \times 1 = 100 \text{ N} $$

Answer: \(F_x = 0\) N, \(F_y = 100\) N

Example 2: A robotic arm moves 2 m vertically (90°). Calculate horizontal and vertical displacements.

Given: length \(= 2\) m, \(\theta = 90^\circ\)

$$ x = 2 \times \cos 90^\circ = 2 \times 0 = 0 \text{ m} $$

$$ y = 2 \times \sin 90^\circ = 2 \times 1 = 2 \text{ m} $$

Answer: \(x = 0\) m, \(y = 2\) m

Example 3: A sensor detects an object at 0° and 90°. If the detection range is 5 m, find the components for both cases.

• For 0°:

$$ x = 5 \times \cos 0^\circ = 5 \times 1 = 5 \text{ m} $$

$$ y = 5 \times \sin 0^\circ = 5 \times 0 = 0 \text{ m} $$

• For 90°:

$$ x = 5 \times \cos 90^\circ = 5 \times 0 = 0 \text{ m} $$

$$ y = 5 \times \sin 90^\circ = 5 \times 1 = 5 \text{ m} $$

Answer: At 0° (5 m, 0 m), at 90° (0 m, 5 m)

Chapter Summary

This chapter covered the application of vector theorems through a detailed study of angles and hyperbolic functions, beginning with the classification of angles into acute, obtuse, reflex, and right angles. It then explored the properties of different types of triangles, including isosceles, equilateral, right-angled, and scalene triangles, which form the basis for trigonometric analysis. The chapter introduced the six fundamental trigonometric ratios: sine, cosine, tangent, cosecant, secant, and cotangent, emphasizing their calculation and use in solving vector problems. Trigonometric identities were examined with proofs, focusing particularly on the Pythagorean identities that simplify complex expressions. The solution of trigonometric equations was addressed to enable the determination of unknown angles and sides in vector contexts. Finally, the chapter concluded with an introduction to hyperbolic functions such as sinh, cosh, cosech, tanh, and sech, highlighting their relevance and application in engineering mathematics. Throughout, the integration of these topics provided a comprehensive foundation for applying vector theorems effectively.