Comprehensive Geometry and Trigonometry Practice Review

The Unit Circle: Principles and Coordinates

Quadrant I ( $0^\circ$ to $90^\circ$ ): * Trigonometric Signs: All functions ($in$, os, $\tan$ , ec, sc, ot) are positive. * $0^\circ$ ( $0$ radians): Point $(1, 0)$ . * $30^\circ$ ( $\frac{\pi}{6}$ radians): Point $(\frac{\sqrt{3}}{2}, \frac{1}{2})$ . * $45^\circ$ ( $\frac{\pi}{4}$ radians): Point $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ . * $60^\circ$ ( $\frac{\pi}{3}$ radians): Point $(\frac{1}{2}, \frac{\sqrt{3}}{2})$ . * $90^\circ$ ( $\frac{\pi}{2}$ radians): Point $(0, 1)$ .
Quadrant II ( $90^\circ$ to $180^\circ$ ): * Trigonometric Signs: in and sc are positive. os, $\tan$ , ec, and ot are negative. * $120^\circ$ ( $\frac{2\pi}{3}$ radians): Point $(-\frac{1}{2}, \frac{\sqrt{3}}{2})$ . * $135^\circ$ ( $\frac{3\pi}{4}$ radians): Point $(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ . * $150^\circ$ ( $\frac{5\pi}{6}$ radians): Point $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$ . * $180^\circ$ ( $\pi$ radians): Point $(-1, 0)$ .
Quadrant III ( $180^\circ$ to $270^\circ$ ): * Trigonometric Signs: $\tan$ and ot are positive. in, os, ec, and sc are negative. * $210^\circ$ ( $\frac{7\pi}{6}$ radians): Point $(-\frac{\sqrt{3}}{2}, -\frac{1}{2})$ . * $225^\circ$ ( $\frac{5\pi}{4}$ radians): Point $(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})$ . * $240^\circ$ ( $\frac{4\pi}{3}$ radians): Point $(-\frac{1}{2}, -\frac{\sqrt{3}}{2})$ . * $270^\circ$ ( $\frac{3\pi}{2}$ radians): Point $(0, -1)$ .
Quadrant IV ( $270^\circ$ to $360^\circ$ ): * Trigonometric Signs: os and ec are positive. in, $\tan$ , sc, and ot are negative. * $300^\circ$ ( $\frac{5\pi}{3}$ radians): Point $(\frac{1}{2}, -\frac{\sqrt{3}}{2})$ . * $315^\circ$ ( $\frac{7\pi}{4}$ radians): Point $(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})$ . * $330^\circ$ ( $\frac{11\pi}{6}$ radians): Point $(\frac{\sqrt{3}}{2}, -\frac{1}{2})$ . * $360^\circ$ ( $2\pi$ radians): Point $(1, 0)$ .

Section 1: Circles (Equations, Area, and Circumference)

Circle Equation Construction: The general form for the equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$ , where $(h, k)$ is the center and $r$ is the radius.
Problem 1: Find the center and radius of $(x - 5)^2 + (y + 2)^2 = 49$ . * Center: $(5, -2)$ * Radius: $\sqrt{49} = 7$
Problem 2: Write the equation of a circle with center $(-3, 0)$ and radius $4$ . * Equation: $(x + 3)^2 + y^2 = 16$
Problem 3: Find the area of a circle with a diameter of $12\,cm$ . * Radius: $6\,cm$ * Area Formula: $A = \pi r^2$ * Calculation: $A = \pi(6)^2 = 36\pi\,cm^2$
Problem 4: Find the circumference of a circle with an area of $64\pi\,units^2$ . * $r^2 = 64 \rightarrow r = 8$ * Circumference formula: $C = 2\pi r$ * Calculation: $C = 2\pi(8) = 16\pi\,units$
Problem 5: A circle has a circumference of $18\pi$ . Find its radius. * $2\pi r = 18\pi \rightarrow r = 9$
Problem 6: Write the equation of a circle centered at the origin that passes through $(0, 6)$ . * Radius is the distance from $(0,0)$ to $(0,6)$ , which is $6$ . * Equation: $x^2 + y^2 = 36$
Problem 7: Find the diameter of a circle whose area is $121\pi\,in^2$ . * $r^2 = 121 \rightarrow r = 11$ * Diameter: $2r = 22\,in$
Problem 8: Identify the center of the circle: $(x + 8)^2 + y^2 = 100$ . * Center: $(-8, 0)$

Section 2: Arcs, Sectors, and Chords

Arc Length and Sector Area Formulas: * Arc Length: $L = \frac{\theta}{360} \times 2\pi r$ (for degrees) or $L = r\theta$ (for radians). * Sector Area: $A = \frac{\theta}{360} \times \pi r^2$ (for degrees) or $A = \frac{1}{2}r^2\theta$ (for radians).
Problem 9: Find the length of an arc with a measure of $60^\circ$ in a circle with radius $10$ . * Calculation: $L = \frac{60}{360} \times 20\pi = \frac{1}{6} \times 20\pi = \frac{10\pi}{3}$
Problem 10: Calculate the area of a sector with a central angle of $120^\circ$ and radius $6$ . * Calculation: $A = \frac{120}{360} \times \pi(6)^2 = \frac{1}{3} \times 36\pi = 12\pi$
Problem 11: If a chord is $8\,units$ from the center of a circle with radius $10$ , how long is the chord? * Use Pythagorean Theorem on the half-chord: $8^2 + x^2 = 10^2 \rightarrow 64 + x^2 = 100 \rightarrow x = 6$ . * Total chord length: $2x = 12\,units$ .
Problem 12: Find the measure of a central angle if its arc length is $4\pi$ and the radius is $8$ . * $L = r\theta \rightarrow 4\pi = 8\theta \rightarrow \theta = \frac{\pi}{2}$ radians (or $90^\circ$ ).
Problem 13: In a circle, two chords intersect. If the segments of one are $4$ and $9$ , and one segment of the other is $6$ , find the missing segment ( $x$ ). * Formula: $a \cdot b = c \cdot d$ * Calculation: $4 \times 9 = 6x \rightarrow 36 = 6x \rightarrow x = 6$
Problem 14: Find the area of a circle where a $90^\circ$ sector has an area of $16\pi$ . * $16\pi = \frac{90}{360} \times A_{circle} = \frac{1}{4} \times A_{circle} \rightarrow A_{circle} = 64\pi$
Problem 15: If an arc measure is $45^\circ$ and the radius is $12$ , find the sector area. * Calculation: $A = \frac{45}{360} \times \pi(12)^2 = \frac{1}{8} \times 144\pi = 18\pi$

Section 3: Tangents and Secants

Lengths of Segments: * Tangent-Secant Theorem: $(\text{Tangent})^2 = (\text{External Secant}) \times (\text{Total Secant})$ * Two-Tangent Theorem: Tangents from the same exterior point to a circle are congruent ( $T_1 = T_2$ ).
Problem 16: A tangent segment and a secant segment are drawn to a circle from an exterior point. If the tangent is $6$ and the external part of the secant is $4$ , find the total length of the secant ( $s$ ). * Calculation: $6^2 = 4 \times s \rightarrow 36 = 4s \rightarrow s = 9$
Problem 17: Two tangents are drawn to a circle from point P. If one tangent is $3x + 5$ and the other is $20$ , solve for $x$ . * $3x + 5 = 20 \rightarrow 3x = 15 \rightarrow x = 5$
Problem 18: Find the measure of an angle formed by two tangents if the intercepted major arc is $240^\circ$ . * Minor arc = $360^\circ - 240^\circ = 120^\circ$ * Angle formula: $\frac{\text{Major Arc} - \text{Minor Arc}}{2} = \frac{240 - 120}{2} = 60^\circ$
Problem 19: A radius is drawn to a point of tangency. What is the angle formed? * The angle is always $90^\circ$ ; radii are perpendicular to tangents at the point of contact.
Problem 20: Find the length of a tangent from a point $13\,units$ from the center of a circle with radius $5$ . * Form a right triangle with the radius and the segment to the center. $5^2 + T^2 = 13^2 \rightarrow 25 + T^2 = 169 \rightarrow T = 12$

Section 4: Inscribed Angles and Shapes

Problem 21: Find the measure of an inscribed angle that intercepts an arc of $110^\circ$ . * Angle = $\frac{1}{2}(\text{Arc}) = \frac{110}{2} = 55^\circ$
Problem 22: If an inscribed angle is $40^\circ$ , what is the measure of its intercepted arc? * Arc = $2 \times (\text{Angle}) = 2 \times 40 = 80^\circ$
Problem 23: An angle is inscribed in a semicircle. What is its degree measure? * The degree measure is always $90^\circ$ .
Problem 24: A quadrilateral is inscribed in a circle. If one angle is $85^\circ$ , find the measure of the opposite angle. * Opposite angles are supplementary: $180 - 85 = 95^\circ$
Problem 25: Find the value of $x$ if two inscribed angles intercept the same arc and are represented by $2x + 10$ and $50^\circ$ . * $2x + 10 = 50 \rightarrow 2x = 40 \rightarrow x = 20$

Section 5: Interior and Exterior Angles of Polygons

Formulas: * Sum of Interior Angles: $(n - 2) \times 180^\circ$ * One Interior Angle (Regular): $\frac{(n - 2) \times 180}{n}$ * Sum of Exterior Angles: Always $360^\circ$ * One Exterior Angle (Regular): $\frac{360}{n}$
Problem 26: Find the sum of the interior angles of a convex heptagon ( $7$ sides). * $(7 - 2) \times 180 = 5 \times 180 = 900^\circ$
Problem 27: What is the measure of one interior angle of a regular octagon ( $8$ sides)? * Total sum: $6 \times 180 = 1080^\circ$ . One angle: $\frac{1080}{8} = 135^\circ$
Problem 28: Find the sum of the exterior angles of a $20$ -gon. * The sum is always $360^\circ$
Problem 29: What is the measure of one exterior angle of a regular decagon ( $10$ sides)? * Calculation: $\frac{360}{10} = 36^\circ$
Problem 30: If the sum of the interior angles is $1800^\circ$ , how many sides does the polygon have? * $(n - 2) \times 180 = 1800 \rightarrow n - 2 = 10 \rightarrow n = 12$
Problem 31: If one interior angle of a regular polygon is $144^\circ$ , how many sides does it have? * Exterior angle: $180 - 144 = 36^\circ$ . Sides: $\frac{360}{36} = 10$
Problem 32: Find the measure of one exterior angle of a regular $15$ -gon. * Calculation: $\frac{360}{15} = 24^\circ$
Problem 33: If an exterior angle is $20^\circ$ , how many sides does the regular polygon have? * Calculation: $\frac{360}{20} = 18$
Problem 34: Find the sum of interior angles for a $12$ -sided polygon. * Calculation: $(12 - 2) \times 180 = 1800^\circ$
Problem 35: Can a regular polygon have an interior angle of $100^\circ$ ? * Exterior angle would be $80^\circ$ . Sides $= \frac{360}{80} = 4.5$ . Since the number of sides must be an integer, no.

Section 6: Properties of Quadrilaterals

Problem 36: In parallelogram $ABCD$ , if $m\angle A = 70^\circ$ , find $m\angle B$ . * Consecutive angles are supplementary: $180 - 70 = 110^\circ$
Problem 37: In a rectangle, the diagonals are $3x - 4$ and $2x + 6$ . Find $x$ . * Diagonals are equal: $3x - 4 = 2x + 6 \rightarrow x = 10$
Problem 38: True or False: Every rhombus is a square. * False. A square must have four right angles.
Problem 39: In a rhombus, the diagonals are $12$ and $16$ . Find the side length. * Diagonals bisect at right angles: use halves ( $6$ and $8$ ) in Pythagorean Theorem: $6^2 + 8^2 = s^2 \rightarrow s = 10$ .
Problem 40: Find the median of a trapezoid with bases of $15$ and $25$ . * Median formula: $\frac{\text{base}_1 + \text{base}_2}{2} = \frac{15 + 25}{2} = 20$
Problem 41: In an isosceles trapezoid, if one base angle is $50^\circ$ , find the other angles. * Angles are $50^\circ, 50^\circ, 130^\circ, 130^\circ$ .
Problem 42: A kite has diagonals of length $10$ and $20$ . Find its area. * Area formula: $\frac{d_1 \times d_2}{2} = \frac{10 \times 20}{2} = 100$
Problem 43: If the diagonals of a quadrilateral bisect each other and are perpendicular, what is the most specific name for it? * A rhombus.
Problem 44: Find the perimeter of a square with a diagonal of $10\sqrt{2}$ . * Side length $s = 10$ . Perimeter: $4s = 40$ .
Problem 45: In parallelogram $JKLM$ , if $JK = 5x - 2$ and $ML = 2x + 10$ , find $JK$ . * Opposite sides are equal: $5x - 2 = 2x + 10 \rightarrow 3x = 12 \rightarrow x = 4$ . * $JK = 5(4) - 2 = 18$ .

Section 7: Mixed Geometry Review

Problem 46: Find the distance between $(1, 2)$ and $(4, 6)$ . * Distance format: $d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2} = 5$
Problem 47: Find the midpoint of a segment with endpoints $(-2, 8)$ and $(6, 0)$ . * Calculation: $(\frac{-2+6}{2}, \frac{8+0}{2}) = (2, 4)$
Problem 48: Two angles are supplementary. One is $3x$ and the other is $x + 20$ . Find $x$ . * $3x + x + 20 = 180 \rightarrow 4x = 160 \rightarrow x = 40$
Problem 49: Find the area of a triangle with base $10$ and height $12$ . * Area: $\frac{1}{2} \times 10 \times 12 = 60$
Problem 50: What is the Pythagorean Theorem? * $a^2 + b^2 = c^2$ for right triangles.
Problem 51: Find the hypotenuse of a right triangle with legs $9$ and $12$ . * Calculation: $9^2 + 12^2 = 81 + 144 = 225 \rightarrow c = 15$
Problem 52: Find the volume of a cylinder with radius $3$ and height $10$ . (Leave in $\pi$ ). * Volume formula: $V = \pi r^2 h$ * Calculation: $V = \pi(3)^2(10) = 90\pi\,units^3$
Problem 53: If two triangles are similar with a scale factor of $2:3$ , what is the ratio of their areas? * Ratio of areas: $(2^2):(3^2) = 4:9$
Problem 54: Find the surface area of a cube with side length $5$ . * Surface Area formula: $SA = 6s^2$ * Calculation: $SA = 6(5)^2 = 150\,units^2$

Part 1: Surface Area and Volume (Advanced Practice)

Problem 1: Find the volume of a rectangular prism with length $8$ , width $3$ , and height $5$ . * Calculation: $8 \times 3 \times 5 = 120$
Problem 2: Find the surface area of a cylinder with radius $3$ and height $10$ . (Use $\pi$ ). * Formula: $2\pi rh + 2\pi r^2$ * Calculation: $2\pi(3)(10) + 2\pi(3)^2 = 60\pi + 18\pi = 78\pi$
Problem 3: Find the volume of a cylinder with radius $4$ and height $7$ . * Calculation: $\pi(4)^2(7) = 112\pi$
Problem 4: Find the volume of a cone with radius $6$ and height $8$ . * Formula: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(36)(8) = 96\pi$
Problem 5: A square pyramid has a base side of $10$ and a height of $12$ . Find its volume. * Formula: $V = \frac{1}{3}Bh = \frac{1}{3}(10^2)(12) = 400$
Problem 6: A square pyramid has a base side of $6$ and a slant height ( $l$ ) of $5$ . Find its surface area. * Formula: $B + \frac{1}{2}Pl = 36 + \frac{1}{2}(24)(5) = 36 + 60 = 96$
Problem 7: Find the volume of a sphere with a radius of $3$ . * Formula: $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(27) = 36\pi$
Problem 8: Find the surface area of a sphere with a radius of $5$ . * Formula: $SA = 4\pi r^2 = 4\pi(25) = 100\pi$
Problem 9: Find the volume of a triangular prism with a base area of $20$ and a height of $15$ . * Calculation: $20 \times 15 = 300$
Problem 10: Find the volume of a cone with a diameter of $10$ and height of $12$ . * Radius: $5$ . Volume: $\frac{1}{3}\pi(25)(12) = 100\pi$

Part 2: Similar Solids

Linear Ratios vs Area vs Volume: * Sides ratio: $a:b$ * Surface Area ratio: $a^2:b^2$ * Volume ratio: $a^3:b^3$
Problem 11: If the ratio of sides of two cubes is $2:5$ , find area ratio. * Ratio: $4:25$
Problem 12: Area ratio is $16:49$ . Find radii ratio. * Ratio: $\sqrt{16}:\sqrt{49} = 4:7$
Problem 13: Side ratio is $3:4$ . Find volume ratio. * Ratio: $3^3:4^3 = 27:64$
Problem 14: Volume ratio is $8:27$ . Find height ratio. * Ratio: $\sqrt[3]{8}:\sqrt[3]{27} = 2:3$
Problem 15: Volumes are $64$ and $512$ . Find surface area ratio. * Side ratio: $\sqrt[3]{64}:\sqrt[3]{512} = 4:8 = 1:2$ . Area ratio: $1^2:2^2 = 1:4$
Problem 16: Heights are $3$ and $9$ . Ratio $1:3$ . Volume ratio $1:27$ . If small volume is $10$ , large volume is $270$ .
Problem 17: Area ratio $9:25 \rightarrow$ Side ratio $3:5 \rightarrow$ Volume ratio $27:125$ . If large volume is $250$ , small is $(27/125) \times 250 = 54$ .
Problem 18: Height ratio $5:15 = 1:3 \rightarrow$ Volume ratio $1:27$ .
Problem 19: Radii ratio $2:6 = 1:3 \rightarrow$ Volume ratio $1:27$ .
Problem 20: Volume ratio $1:125 \rightarrow$ Side ratio $1:5 \rightarrow$ Area ratio $1:25$ .

Part 3: Regular Polygons

Problem 21: Find the area of a regular hexagon with side length $8$ . * Area: $\frac{3\sqrt{3}}{2}s^2 = \frac{3\sqrt{3}}{2}(64) = 96\sqrt{3}$
Problem 22: Find the apothem of a square with side length $12$ . * Apothem: $\frac{1}{2}s = 6$
Problem 23: Find area of regular pentagon: apothem $4$ , side $5.8$ . * Area: \frac{1}{2}n a = \frac{1}{2}(5)(5.8)(4) = 58
Problem 24: Area of equilateral triangle with side $10$ . * Formula: $\frac{s^2\sqrt{3}}{4} = \frac{100\sqrt{3}}{4} = 25\sqrt{3}$
Problem 25: Regular octagon area: apothem $10$ , perimeter $66$ . * Area: $\frac{1}{2}Pa = \frac{1}{2}(66)(10) = 330$
Problem 26: Central angle of regular decagon ( $10$ sides). * Calculation: $\frac{360}{10} = 36^\circ$
Problem 27: Regular hexagon has apothem $5\sqrt{3}$ . Find side length. * In a hexagon, $a = s\frac{\sqrt{3}}{2} \rightarrow 5\sqrt{3} = s\frac{\sqrt{3}}{2} \rightarrow s = 10$
Problem 28: Area of regular hexagon with apothem $6$ . * $6 = s\frac{\sqrt{3}}{2} \rightarrow s = \frac{12}{\sqrt{3}} = 4\sqrt{3}$ . Area: $\frac{1}{2}(6 \times 4\sqrt{3})(6) = 72\sqrt{3}$
Problem 29: Perimeter of square with apothem $5$ . * Side $= 10$ . Perimeter $= 40$ .
Problem 30: Area of regular polygon: $n = 12$ , side $s = 4$ , apothem $a = 7.5$ . * Perimeter $= 12 \times 4 = 48$ . Area: $\frac{1}{2}(48)(7.5) = 180$

Part 4: Circle Angles (Vertex Inside, Outside, and On)

Problem 33: Two chords intersect inside. Arcs are $40^\circ$ and $80^\circ$ . * Angle: $\frac{40 + 80}{2} = 60^\circ$
Problem 34: Two secants outside. Arcs are $100^\circ$ and $30^\circ$ . * Angle: $\frac{100 - 30}{2} = 35^\circ$
Problem 35: Tangent and secant outside. Arcs are $120^\circ$ and $50^\circ$ . * Angle: $\frac{120 - 50}{2} = 35^\circ$
Problem 36: Angle formed by tangent and chord intercepts arc of $140^\circ$ . * Angle: $\frac{140}{2} = 70^\circ$
Problem 38: Chords intersect inside. Angle is $100^\circ$ , one arc is $130^\circ$ . Find other arc ( $x$ ). * $100 = \frac{130 + x}{2} \rightarrow 200 = 130 + x \rightarrow x = 70^\circ$
Problem 39: Angle outside is $40^\circ$ . Larger arc is $110^\circ$ . Find smaller arc ( $x$ ). * $40 = \frac{110 - x}{2} \rightarrow 80 = 110 - x \rightarrow x = 30^\circ$
Problem 42: Two tangents from external point create major arc of $250^\circ$ . * Minor arc: $360 - 250 = 110^\circ$ . Intersection angle: $\frac{250 - 110}{2} = 70^\circ$
Problem 45: Two secants outside. Arcs are $150^\circ$ and $x$ . Angle is $50^\circ$ . * $50 = \frac{150 - x}{2} \rightarrow 100 = 150 - x \rightarrow x = 50^\circ$
Problem 50: Two tangents meet at $60^\circ$ angle. Find minor arc ( $x$ ). * $60 = \frac{(360 - x) - x}{2} \rightarrow 120 = 360 - 2x \rightarrow 2x = 240 \rightarrow x = 120^\circ$ (Wait: $2x = 240 \rightarrow x = 120$ . Correction: minor arc is $120^\circ$

Circle Geometry Practice: Chords and Intersections

Problem 1: Radius $= 13$ , chord distance from center $= 5$ . * Logic: Half-chord is $b^2 = 13^2 - 5^2 = 144 \rightarrow b = 12$ . Chord length $= 24$ .
Problem 2: Chord length $= 24$ , distance from center $= 9$ . * Radius: $r^2 = 12^2 + 9^2 = 144 + 81 = 225 \rightarrow r = 15$ .
Problem 5: Tangent $= 12$ , External secant segment $= 8$ . Total secant length ( $s$ ). * $12^2 = 8(s) \rightarrow 144 = 8s \rightarrow s = 18$ .
Problem 6: Tangent $x$ , External secant $4$ , Internal secant $12$ ( $total = 16$ ). * $x^2 = 4(16) \rightarrow x^2 = 64 \rightarrow x = 8$ .
Problem 7: Secant 1 ( $E=5, I=7 \rightarrow T=12$ ). Secant 2 ( $E=4, I=x \rightarrow T=4+x$ ). * $5(12) = 4(4+x) \rightarrow 60 = 16 + 4x \rightarrow 44 = 4x \rightarrow x = 11$ .

Part 1: Solving Triangles (Laws of Sines & Cosines)

Problem 1: In $\triangle ABC$ , $A = 40^\circ$ , $B = 60^\circ$ , $c = 20$ . Find $a$ . * $C = 180 - (40+60) = 80^\circ$ . Law of Sines: $\frac{a}{\sin(40^\circ)} = \frac{20}{\sin(80^\circ)}$ .
Problem 2: Sides $a=10, b=15, c=20$ . Find $C$ . * Law of Cosines: $20^2 = 10^2 + 15^2 - 2(10)(15)\cos(C) \rightarrow 400 = 100 + 225 - 300\cos(C)$ .
Problem 3: $a=12, b=18, C=45^\circ$ . Area? * Area: $\frac{1}{2}ab\sin(C) = \frac{1}{2}(12)(18)\sin(45^\circ) = 108(\frac{\sqrt{2}}{2}) = 54\sqrt{2}$ .
Problem 7: Heron's Formula for sides $8, 15, 17$ . * Semi-perimeter $s = \frac{8+15+17}{2} = 20$ . Area $= \sqrt{20(20-8)(20-15)(20-17)} = \sqrt{20 \cdot 12 \cdot 5 \cdot 3} = \sqrt{3600} = 60$ .

Trigonometry: Unit Circle & Basic Values

Problem 11: Evaluate $\sin(150^\circ)$ . * Value: $\frac{1}{2}$
Problem 12: Evaluate $\cos(\frac{4\pi}{3})$ . * Value: $-\frac{1}{2}$
Problem 13: Evaluate $\tan(225^\circ)$ . * Value: $1$
Problem 14: Evaluate $\csc(\frac{7\pi}{4})$ . * Ratio: $\frac{1}{-\frac{\sqrt{2}}{2}} = -\sqrt{2}$
Problem 31: If \sin(\theta) > 0 and \cos(\theta) < 0, quadrant? * Quadrant II.
Problem 32: If \tan(\theta) < 0 and \cos(\theta) > 0, quadrant? * Quadrant IV.
Problem 36: Simplify $\sin^2(\theta) + \cos^2(\theta)$ . * Result: $1$
Problem 37: Simplify $1 + \tan^2(\theta)$ . * Result: $\sec^2(\theta)$

Vector Applications Practice

Problem 1: Hiker walks $4\,miles$ at bearing $100^\circ$ , then $7\,miles$ at bearing $250^\circ$ .
Problem 4: Plane airspeed $500\,mph$ at bearing $N60^\circ E$ . Wind from West at $40\,mph$ . * Plane vector: $V_p = \langle 500\cos(30^\circ), 500\sin(30^\circ) \rangle$ . * Wind vector: $V_w = \langle 40, 0 \rangle$ .
Problem 5: Boat velocity $20\,mph$ due North. Current $8\,mph$ due East. * Resultant speed: $\sqrt{20^2 + 8^2} = \sqrt{464} \approx 21.5\,mph$ .
Problem 7: $500\,lb$ weight hanging from two cables ( $30^\circ$ and $45^\circ$ angles). * System of equations for tension: $T_{1}\cos(30^\circ) = T_{2}\cos(45^\circ)$ and $T_{1}\sin(30^\circ) + T_{2}\sin(45^\circ) = 500$ .