Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Euler angles wikipedia , lookup
Golden ratio wikipedia , lookup
Perceived visual angle wikipedia , lookup
Reuleaux triangle wikipedia , lookup
Rational trigonometry wikipedia , lookup
History of trigonometry wikipedia , lookup
Trigonometric functions wikipedia , lookup
Incircle and excircles of a triangle wikipedia , lookup
Euclidean geometry wikipedia , lookup
MATH FINAL Please note that throughout this file << means angle O B Circles and angles inside circles: A Arc: smallest arc between two points. An arc is measured in radians. A ɣ <<AB (arc) = αr α length = arc<<AB r 360o = 2; 1o = /180o ɣ 2ɣ α B O r Circumference (C) of circle = 2R Central Angle and Inscribed Angle: B O 2δ r β δ r Being that O is the center of the circumference: δ δ α = central angle; α length = arc<<AC; α = 2β C δ - the proof: if a line is drawn from point B to point O (forming a radius) then β is divided into ɣ and δ. The two triangles formed are both isosceles since two of their sides are radii (which are always congruent.) The external angle O equals either 2ɣ or 2δ the sum of 2ɣ + 2δ = α thereby we know that α = 2β β = inscribed angle; β = α/2; β = (arc<<AC)/2 If there are a few inscribed angles all of whose endpoints lie on the two endpoint of the same intercepted arc than they are congruent <<A and <<B in this picture are P congruent to each other β In every quadrilateral that can be inscribed the opposite angles are supplementary (the sum of the two angles = 180o) 360 -2β O Interior Eccentric Angle: Exterior Eccentric Angle: A 2β B A A <<AB + <<CD <<AB - <<CD 180 -β P C α = ----------------β = ----------------C α 2 2 Q B D B β Circumscribed Angle: Segment Angle: D <<ACB - <<AB α = ----------------------2 <<AB β = -----2 A C 180-2(90 -β) = 2β O O α β B 2β r 90 -β r 2β B P A Thales Theorem: if A,B, and C are points on a circle where the line AC is a diameter 90 -β of the circle, then the angle ABC is a right angle α Proof: AC=CB=CD=radius, thereby triangle OBC is isosceles, and triangle AOB is also isosceles. Triangle ABC = α +(α+ β) + β = 180o 2α + 2β = 180o α +β = 90o αβ β Basic Trigonometry The following is a simple chart with the trig that can help us on the final In order to remember how to use sine, cosine, and tangent simply remember the acronym SOHCAHTOA – which is deciphered: Sine is Opposite over Hypotenuse; Cosine is Adjacent over Hypotenuse; Tangent is Opposite over Adjacent. Congruence & Similarity of Triangles Triangles can be congruent by: SSS (side-side-side), ASA (angle-side-angle), SAS (side-angle-side), SAA (side-angle-angle), and HL (height-leg – if there is a right angle where the height of the triangles and the leg attached to the 90o angle are congruent then the two triangles are congruent) δ Note: triangles can not be congruent by AAA (angle-angle-angle) or by SSA (side-side-angle – unless the angle is a right angle) Triangles can be similar (all angles congruent and side a proportional to each other) by: AA (angle-angle – if two angles are congruent the third will also be congruent), SAS (side-angle-side), SSS (side-side-side), HL (height-leg – if there is a right angle where the height of the triangles and the leg attached to the 90o angle, where the corresponding sides have identical ratios), Note: SSA (side-side-angle – unless the angle is a right angle) The relationship between a point P inside a circumference P is internal: P is external: A (PA)(PB) = (PC)(PD) (PA)(PB)=(PC)(PD) C P D B Proof: <<APD is congruent to <<CPB (opposite by same vertex) <<DAB is congruent to <<BCD (inscribed angle by arc DB) Triangle APD is thereby similar to triangle CPB by AA So the ratio would be: PA = PC PD PB This means that (PA)(PB) = (PC)(PD) B A P C D Proof: <<BPD is congruent to <<DPB (same angle) <<ADC is congruent to <<CBA (inscribed angle by arc AC) Triangle BCP is thereby similar to triangle DAP by AA So the ratio would be: PA = PC PD PB This means that (PA)(PB) = (PC)(PD) (PM)2 = (PN)2 PM = PN (PA)(PB) = (PM)2 M M P O A P B N Proof: <<MPB is congruent to <<MPA (same angle) <<PMA is congruent to <<MBA (PMA is a segment angle and MBA is an inscribed angle both of the same arc MA) Triangle PMB is thereby similar to triangle PAM by AA So the ratio would be: PM = PB PA PM This means that (PA)(PB) =A(PC)(PD) D A+C=B=D C Proof: <<OMP =90o definition of tangent line <<ONP =90o definition of tangent line <<OMP = <<ONP if two angles are rt. << then they’re congruent OM = ON radii of same circle are congruent OP =OP reflexive property Triangle POM is congruent to triangle PON SSA PM = PN corresponding parts of congruent triangles are congruent B Cevians of Triangles: line segments where a vertex forms one of the endpoints and the opposite side is another endpoint Median – the line segment going from one vertex to the midpoint on the opposite side Centroid or center of gravity – the C.G. of a triangle is the point of intersection of the medians of this triangle [property: the C.G. divides the median where from C.G. to vertex is twice the length as from C.G. to opposite side – thus forming a 2:1 ratio.] Altitude – the line segment going from one vertex to the opposite side making a 90o angle [an acute triangle has 3 altitudes, a right triangles has 1 altitude that is not part of the triangle (the other two altitudes would be the two legs of the triangle) an obtuse triangle as 1 altitude.] Orthocenter – the orthocenter of a triangle is a point of intersection of all altitudes [the orthocenter is only interior to the triangle when its acute] Internal bisectors – segments that split in half the angle of the vertex and has the opposite side as its endpoint [every triangle has 3 internal bisectors] Incenter – the incenter of a triangle is the point of intersection of all internal bisectors [the incenter is always interior to the triangle, and it is also the center of a circumference inscribed in a triangle] Remarks about the points of intersection: If triangle ABC is isosceles, then the three points of intersection are collinear If triangle ABC is equilateral then the three points coincide Areas Parallelogram: A = bh Square: A=l2 b l h l Rectangle: A = (width)(height) Trapezoid: Triangle: A = bh/2 Rhombus: A = ½(d1d2) h b b b radius The radius of a regular polygon is the radius of a circumscribed circumference – it bisects the angle of the polygon apothem The apothem of a regular polygon is the radius of the inscribed circumference All apothems of regular polygons are congruent Only regular polygons have apothems An apothem forms a right angle with a side The area of a regular polygon is A= ½(apothem)(perimeter) The area of a regular hexagon is A = ½(s2(route3)) To compare ratio of areas: Since the ration of the sides is x/y and the height changes the ratio of the smaller x triangle to the bigger triangle = x2/y2 the area of the trapezoid formed would be x2/(y2 - x2) y In a triangle the mid-line = ½ the length of the base In a trapezoid the mid-line =the average between the big base and small base ½(big+small) Congruent Cevians Theorems: 1. corresponding medians of congruent triangles are congruent triangle ABC is congruent to triangle DEF given A AB is congruent to ED CPCTC <<A is congruent to <<D CPCTC AR is congruent to RC definition of median DS is congruent to SF definition of median AR is congruent to DS subtraction Triangle ABR is congruent to triangle DES SAS BR is congruent to ES CPCTC D R S B E C C F