Jenna LangMr. Rice / Sec: 5 Adv. Final Study Guide Project (7,10,16)15 June 2015

Chapter 7: Polygons

7.1 Triangle Application Theorems-the sum of the measures of the three angles of a triangle is 180-the measure of the exterior angle of a triangle is equal to the sum of the measures of the remote exterior angles

-Midline Theorem: a segment joining the midpoints of two sides of a triangle is parallel to the third side, and its length is one-half the length of the third side

7.2 Two Proof-Oriented Triangle Theorems-No-Choice Theorem: if two angles of one triangle are congruent to two angles of another triangle, the third angles are also congruent-AAS: If two angles and a non-included side of one triangle are congruent to two angles and a non-included side of another triangle, the triangles are congruent

7.3 Formulas Involving Polygons-the sum Si of the measures of the angles of a polygon with n sides is

Si = (n-2)180-the sum of one exterior angles from each vertex of any polygon is 360-the number d of diagonals from a polygon of n sides is

d = n(n−3)2

7.4 Regular Polygons

-regular polygons are both equilateral and equiangular-the measure E of each exterior angle of an equiangular polygon of n sides is

E = 360n

-ex. A Hexagon (6 sides) has 60∘ exterior angles-the measure of each of a regular polygon’s interior angles is 180-exterior angle measure-ex. The hexagon’s interior angles each measure 120∘ because 180-60=120

Chapter 10: Circles

10.1 The Circle

-things to know about circles:

concentric circles / circle vocabulary (radius, diameter, chord, secant, tangent)

-Areacircle = π r2 (r❑= radius)-Circumference (basically the perimeter) = π d (d❑= diameter=2r❑)

-circle theorems

-if a radius is ⊥to a chord, then it bisects the chord (the opposite works, too)-the ⊥bisector of a chord always passes through the center of the circle

10.2 Congruent Circles-if two chords are equidistant from the center of the circle, they are congruent (opposite works, too)

10.3 Arcs of a Circle-arc: 2 points on a circle and all points on the circle needed to connect the points by a

single path-the center of an arc is the center of the circle of which the arc is a part-minor arc: an arc whose points are on or between the sides of a central angle-major arc: are whose points are on or outside of a central angle

-central angle: angle whose vertex is at the center of a circle-semicircle: half a circle, the endpoints are the endpoints of a diameter-measure of an arc: measure of the angle that intercepts the arc

-in order to for two arcs to be congruent, they need to have both the same measure and be parts of the same or congruent circles

-congruent segments⇔congruent arcs⇔ congruent central angles

10.4 Secants and Tangents-secant: line intersecting a circle at exactly 2 points (see 10.1 diagram)

-a secant segment is the part of a secant line that joins a point outside the circle to the farther intersection point of the secant and the circle

-tangent: line intersecting a circle at exactly 1 point(called the point of tangency/contact)-a tangent line is perpendicular to the radius drawn to the point of contact-a tangent segment is the part of a tangent line between the point of contact and a point outside the circle-Two-Tangent Theorem: if 2 tangent segments are drawn to a circle from an external point, then those segments are congruent -tangent circles: circles that intersect each other at exactly 1 point-internal vs external tangent circles

10.5 Angles Related to a Circle

10.6 More Angle-Arc Theorems-if 2 inscribed or tangent-chord angles intercept the same/congruent arcs, then they are congruent-an angle inscribed in a semicircle is a right angle

-the sum of the measures of a tangent-tangent angle and its minor arc is 180∘

10.7 Inscribed and Circumscribed Polygons

(from the point of view of the triangle)-the center of the circle inscribed in a polygon is the incenter of the polygon-the center of a circle circumscribed about a polygon is the circumcenter of the polygon-if a quadrilateral is inscribed in a circle, its opposite angles are supplementary-if a parallelogram is inscribed in a circle, it is a rectangle

10.8 The Power Theorems-Chord-Chord Power Theorem: if 2 chords of a circle intersect inside the circle, the product of the measures of the segments of one chord is equal to the products of the measures of the segments of the other chord-Tangent-Secant Power Theorem: if a tangent segment and a secant segment are drawn from an external point to a circle, then the square of the measure of the tangent segment is equal to the product of the measures of the entire secant segment and its external part -Secant-Secant Power Theorem: if 2 secant segments are drawn from an external point to a circle, then the product of the measures of 1 secant segment and its external part if equal to the product of the measures of the other secant segment and its external part

Chord-Chord Tangent-Secant Secant-Secant

-ex. Find x

122=x(x+18) 0=x2+18x-144 0=(x-6)(x+24) x=-24, 6 (but a distance can’t be negative so 24 can’t

work) Answer: x=6

10.9 Circumference and Arc Length-Circumference = π d (d❑= diameter)-Arc Length: linear measurement similar to the length of a line segment (expressed in

units) = anglemeasureof arc

360 (π d)

Chapter 16: Inverses, Polar Coordinates, Vectors

16.1 Inverse Functions, Principal Values-to find the inverse of a function, simply change the x and y values in the coordinates-sin❑−1 x = arcsin x❑-the inverse functions of the 6 basic trig functions are not actually functions (they don’t

pass the vertical line test)- restrict the range to make them functions

16.2 Equations Involving Circular and Trigonometric Functions-the ones that look like this sinx + cscx = 2-to solve them, either graph them on your calculator or factor them (if possible)

-make sure you look to see if the question asks for all solutions of just the solutions within a specific interval (usually 0≤x≤360∘ or 2π)

16.3 Polar Coordinates, Polar Graphs-the “origin” is called “The Pole”-(r, θ) r= the length of the radius θ= the angle formed between the terminal side and

the polar axis (the positive x axis) moving counter-clockwise on the unit circle-if r is negative, move |r| in the opposite direction-if θis negative, move clockwise-any point of the polar graph can be named in an infinite number of ways-polar and Cartesian (regular) coordinates are related

x = rcosθ y = rsinθ r = ±√❑

16.4 Graphs of Complex Numbers-(x,y) becomes x + yι - the graph is called an Argand diagram-polar/trigonometric form = r(cosθ+ ιsinθ) r = |z| where z❑is the amplitude -if z1= r1(cosθ1+ ιsinθ1) and z2= r2(cosθ2+ ιsinθ2) then:

z1z2= r1r2[cos(θ1+ θ2) + ιsin(θ1+ θ2)]z1/¿z2= r1/r2[cos(θ1- θ2) + ιsin(θ1- θ2)]-these can be proved using the sum and difference identities

16.5 De Moivre’s Theorem-if z❑= r(cosθ+ ιsinθ) then zn=¿rn(cosnθ+ ιsinnθ) and

n√❑ z❑= n√r ¿cos θ+k∗360

n + ιsin θ+k∗360

n ) where k❑is 0, 1, 2, 3, ..., n❑-1

-ex. See practice problem #9.

16.6 Vectors in the Plane -vector: a quantity specified by both a magnitude and direction (ex. velocity, force)

-resultant: sum of two vectors geometrically

-dot product: if v = ai+bj and u = ci+dj, then u∙v = ac+bd = ||u||||v||cosθ(by the way, if θ= 90∘, then u∙v = 0) ||u||= magnitude=distance

-difference of two vectors: add one vector to the negated second vector

16.7 Applications of Vectors-vectors are graphed on a compass plane, where N=0∘, E=90∘, S=180∘, W=270∘-to solve these word problems, draw a graph of the givens and what you have to find, then use the Law of Cosines, Law of Sines, SOHCAHTOA, or another method to solve -ex. See practice problem #10.PRACTICE PROBLEMS

1. Given: angle ABE angle ACD FD FE Prove: ΔABC is isosceles

2. A regular polygon has 20 diagonals and a perimeter of 80m. What is the length of the shortest diagonal and the acute angles this diagonal forms with the adjacent sides of the polygon?

3. Which polygon’s apothem forms a 36∘angle with a line connecting the polygon’s center and a vertex?

4. Given: HB HF, circle with center H, angles HBC and HFE are right

Prove: DC DE

5. Given: circle with center C, arc BG arc DF

Prove: AG EF

6.Two logs are being held together with a rope. The logs have radii of 2 m and 5 m. How long does the rope have to be?

7. Given: Diagram as shownFind: angles BCD, AGD, ABD

8. Solve for x. (all solutions)

√❑sin2xcosx+sinx2 = sinxcosx+(√❑sin2x)/2

9. Given: u: (10, 30∘) v: (5, 45∘) Find: A) uv

B) u/v

C) u3

D) 3√v

E) ||u+v||

F) ||u−v||

G) u and v in component form

H) graph u and v

10. A boat sails 20 mph on a bearing of 10∘ with an ocean current of 5 mph from the west. What is the boat’s true course?

ANSWERS TO PRACTICE PROBLEMS

1. Statements Reasons

1. angle ABE angle ACD, FD FE 1. Given

2. angles DBE and ECD are 2. Supplements of angles are

3. angles BFD and CFE and 3. Vertical angles are

4. ΔBFD ΔCFE 4. AAS (2, 3, 1)

5. BF CF 5. CPCTC

6. angles CBF and BCF are 6. sides =

7. angles DBC and ECB are 7. angles + angles = angles

8. angles ABC and ACB are 8. Supplements of angles are

9. AB AC 9. angles = legs10. ΔABC is isosceles with base BC 10. Def. of isosceles Δ

2. Use the formula for the number of diagonals in any regular polygon:

20=x(x−3)2

where x=number of sides of the polygon

x=8 Find the measure of each exterior angle:

3608 = 45∘

Find the measure of each interior angle:180-45=135∘

Find the length of each side:80÷8=10m

Use Law of Cosines to find x (length of shortest diagonal)

x2=102+102-2(10)(10)-cos135∘x=18.48

Find measure of angles b and c, they are congruent because their legs are congruent180 (degrees in every Δ)=135+b+cb and c each measure 22.5 ∘

Answer: The shortest diagonal is 18.48 m and the acute angle between that diagonal and the adjacent side measure 22.5∘

3.

2a=2(36)=72∘360÷72=5A polygon with 5 sides is a pentagon.Answer: A pentagon

4. Statements Reasons

1. HB HF, circle with center H 1. Given angles HBC and HFE are right 2. Draw HC, HE, HD, CE 2. Between any 2 points exists a line

3. HC HE 3. All radii of the same circle are

4. ΔHBC ΔHFE 4. HL Postulate (1, 3, 1)

5. BC FE 5. CPCTC 6. Name the intersection of CE and HD point J 6. Any point can be named

7. HJ HJ, HD HD 7. Reflexive

8. angles HCJ and HEJ are 8. legs = angles

9. ΔHCJ and Δ HEJ are 9. SAS (3, 8, 7)

10. angles HCJ and EHJ are 10. CPCTC

11. angles BHD and FHD are 11. angles + angles = angles

12. ΔBHD ΔFHD 12. SAS (1, 11, 7)

13. BD FD 13. CPCTC

14. DC DE 14. segments- segments =

segments

5. Statements Reasons

1. circle with center C, arc BG arc DF 1. Given 2. Draw CG and CF 2. Between any 2 points exists a line

3. CG CF 3. All radii of the same circle are

4. angles CGF and CFG are 4. legs = angles

5. angles CGA and CFE are 5. Supplements of angles are

6. angles GCB and FCE are 6. arcs = central angles

7. ΔCGA ΔCFE 7. ASA (5, 3, 6)

8. AG EF 8. CPCTC

6.

Red portion: cosa=(3/7) a=64.6∘ 2a=129.2∘ 360-a29.2∘ = 230.75∘

10π(230.75/360)Green portion: sinb=(3/7) b=25.38∘ 2b= 50.75∘ 2b+180=230.75 360-230.75= 129.25

4π(129.25/360)Find x: x2+32=72 x=6.3 Total rope length: 10π(230.75/360)+4π(129.25/360)+2(6.3) = 19.17 mAnswer: 19.17 m

7. angle BCD measures 70∘, angle AGD measures 150∘, angle ABD measures 75∘

8. √❑sin2xcosx +sinx2 =sinxcosx +(√❑sin2x)/2

0=√❑sin2xcosx+sinx2 - sinxcosx - (√❑sin2x)/2

0= (sinxcosx-sinx2 )(√❑sinx-1)

0= (sinx)(cosx-12)(√❑sinx-1)

sinx=0 or cosx-12=0 or √❑sinx-1=0

x=0∘+ 360k, 45∘+360k, 60∘+360k, 135∘+ 360k, 300∘+ 360k, where k is an integer

9. Given: u: (10, 30∘) v: (5, 45∘)

A) uvz1z2=r1r2[cos(θ1+ θ2) + ιsin(θ1+ θ2)]uv=50(cos75∘ + ιsin75∘ )uv=12.9 + 3ι

B) u/v z1/¿z2= r1/r2[cos(θ1- θ2) + ιsin(θ1- θ2)]u/v= 2(cos(-15∘)-ιsin(-15∘))u/v= 1.9+0.3ι

C) u3

zn=¿rn(cosnθ+ ιsinnθ) u3=103(cos90∘+ιsin90∘)u3=1000ι

D) 3√vn√❑ z❑= n√r ¿cos

θ+k∗360n + ιsin

θ+k∗360n ) where k❑is 0, 1, 2, 3, ..., n❑-1

3√❑v= 5❑1/3 ¿cos 45+0∗360

3 + ιsin 45+0∗360

3 ) or 5❑1/3 ¿cos 45+1∗360

3 + ιsin

45+1∗3603 ) or

5❑1/3 ¿cos 45+2∗360

3 + ιsin 45+2∗360

3 )3√v= 1.7+0.4ι or -1.2 +1.2ι or 0.8-0.5ι

E) ||u+v|| (add the two component forms)

(10√❑+ 5√❑)/2, (10+5√❑)/2 F) ||u−v||

(subtract the two component forms)

(10√❑- 5√❑)/2, (10-5√❑)/2

G) u and v in component form

u: 5√❑, 5

v: (5√❑)/2, (5√❑)/2 H) graph u and v

10.

-Draw a second NESW axis where the boat and current lines meet. This NS line is parallel to the original NS line so angle a ≈c and angle c=10∘. -Use Law of Cosines: x2=202+52-2(20)(5)cos100∘ x=21.4

Use Law of Sines: sinb5 =

sin 100∘x b=13.3∘

Total bearing of boat on its true course = b+10∘= 23.3∘Answer: The boat’s true course is 20.4mph at a bearing of 23.3∘