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Similar Triangles B A b a Compare Like Ratios to Like Ratios c Examples: Large A B Left = or Small a b Right C corresponding angles match Slope Grade Pitch Gradient Tangent A a = etc. B b imbedded triangles are similar when P || Q m= rise m= run 45°-45°-90° Triangles y x reflected triangles are similar when P || Q x2 + y2 = z2 z 1 + m2 x= a + b = 90° 30°-60°-90° Triangles Heron's Formula b a zm 1 + m2 y= A s= c A= a c n=sides Q Δy = y2 − y1 d B Δz = z2 − z1 total degrees = (n–2) 180° D2 = Δx2 + Δy2+ Δz2 A = π R2 a As rc R A Δx = x2 − x1 e C s(s–a)(s–b)(s–c) P 3D Pythagorean Theorem b a + b +c 2 D=2R D C=πD C=2πR R θ π R2 360° θ arc = 2πR 360° AS = B1 B2 + (B1 + B2) A= H 2 A= |[ slope = m = D= b B ] – 2 y1 y2 y3 y4 y1 Parallelograms x1 y2 x2 y3 x3 y4 x4 y5 [ ] + | A=BH a + b = 180° x1 + x2 y1 + y2 , ) 2 2 Δx2 + Δy2 b B 4 π R3 3 rise Δy y2 – y1 = = run Δx x2 – x1 midpoint = ( Volume of Frustums x1 x2 x3 x4 x1 x2 y1 x3 y2 x4 y3 x1 y4 H V= Surface Area = 4 π R2 Polygon Area Trapezoids When A || B then <P = <Q (B2 + B b + b2) H V= 3 R r H V= πH 2 (r + Rr + R2) 3 Solids with common cross-sections (such as a barrel) all have similar volumes V = π R2 H V = (ABase)H H height is perpendicular to the base Solids with similar cross-sections (such as a cone) all have similar volumes V= R (ABase)H 3 R Find m: m = Exponents am an = am+n Factor 109 106 103 102 101 10–1 10–2 10–3 10–6 10–9 Example 5 GigaBytes 6 Megahertz 2 kilometers 3 hectograms 5 dekaliters 3 decigrams 2 centimeters 5 milligrams 4 micrograms 5 nanoseconds Mitered Triangles a = am–n an a0 = 1 rise y2 – y1 = run x2 – x1 if the y-intercept is between integers or off the grid b should be computed using b = y0 – m x0 where (x0, y0) is any known point Once m & b are found: y = m x + b Solving Linear Equations 1) 2) 3) 4) 5) 6) 7) 8) Remove Fractions: mult through by LCD cancel fractions Remove ( ): distributive rule; beware of negatives! Combine Like Terms Shift variable terms to one side (add/sub) Shift all other terms to other side Write as (coefficient) × (variable) Divide out the Coefficient Check the Answer Quadratic Formula ax2 + bx + c = 0; m x= -b± b2 − 4ac 2a Distributive Rule 1 a–n = n a a(b ± c) = ab ± ac (a b)n = (an) (bn) (b ± c)a = ba ± ca b±c b c = ± a a a a n an ( b ) = bn a = a1/2 = a0.5 ab = a = b a π R2 H 3 Find b: b = y-intercept (i.e. x = 0) To Graph ax + by = c Plot the intercepts set x = 0, solve for y and plot (0, y) set y = 0, solve for x and plot (x, 0 Draw graph Symbol G M k h da d c m µ n V= To Find a Line's Equation To Graph y= mx + b Plot (0, b) the y-intercept Plot a second point using m by shifting horizontally by run and shifting vertically by rise. Draw graph Prefix Giga Mega Kilo Hecto Deka Deci Centi Milli Micro Nano Cone H height is perpendicular to the base Lines Surface Area = 2πRH + 2πR2 b a b Factoring ax + bx = (a + b)x ax + bx − cx = (a + b − c)x Summary of Trig Relationships y Non-trig Relationships using Slope, m y m= x2 + y2 = r2 x x= r 1 + m2 y= rm 1 + m2 (x, y)rect (r, )polar y θ° a= 2π r (θ in degrees) 360° a = r θ (θ in radians) r Arc Length and Velocity 2πrad = 360° v = r ω (θ in radians) e.g. (ft/sec = ft × rad/sec) a y x x x Standard Diagram Trig Relationships y opp sin θ = = r hyp x adj cos θ = = r hyp y opp tan θ = = x adj Given a right triangle when θ is known use these trig relationships to find a missing side x = r cos θ y = r sin θ θ + α = 90° = π/2rad tan θ = m θ = tan-1 m sin θ = cos α Given a right triangle when θ is not known use these trig relationships to find an angle 45° = π/4 180° = π opp x adj y opp ( r ) = sin–1 ( hyp ) θ = cos–1 ( r ) = cos–1 ( hyp ) θ = tan–1 ( x ) = tan–1 ( adj ) 60° = π/6 sin2 θ + cos2 θ = 1 cos θ = sin α y θ = sin–1 30° = π/12 sin θ = tan θ cos θ Some Examples that do NOT require Trigonometry Not every triangle requires trigonometry. Don't forget to use similar triangles where possible. Don’t forget there are 180° in every triangle. Two Angles Match Imbedded Triangles, P||Q Reflected Triangles, P||Q Use Similar Triangles Use Similar Triangles Use 180° in triangles Use Similar Triangles Imbedded Right Triangles Both legs known, slope missing Mitered Right Triangles Only right triangles can use the Pythagorean Thm. Slope only makes sense with a vertical vs. horizontal relationship. Use Similar Triangles Use Pythagorean Thm. Use m = rise run Use Similar Triangles Use Pythagorean Thm. A right triangle with 2 sides known, one side missing Use Pythagorean Thm. Slope and diagonal known, legs missing Slope and one leg known, other leg missing rise Use m = run Use x= z ; y= 1 + m2 zm 1 + m2 Examples that do require Trigonometry The tangent is used when There is a right triangle The parts of interest are: Both legs known, angles missing one angle and one leg known, other leg missing Use tan θ = opp/adj Use θ = tan-1 (opp/adj) an angle, both legs, the slope Slope known, Standard angle missing Standard angle known, slope missing Use tan θ = m -1 Use θ = tan m The sine is used when There is a right triangle The parts of interest are: an angle, its opposite leg, the diagonal Angle, opposite leg known, hypotenuse missing Angle, hypotenuse known, leg opposite angle missing Hypotenuse, leg known; angle opposite leg missing Use sin θ = opp/hyp Use sin θ = opp/hyp Use θ = sin-1 (opp/hyp) Angle, adjacent leg known, hypotenuse missing Angle, hypotenuse known, adjacent leg missing Known sides bracketing a missing angle Use cos θ = adj/hyp Use cos θ = adj/hyp Use θ = cos-1 (adj/hyp) The cosine is used when There is a right triangle The parts of interest are: an angle, its adjacent leg, the diagonal When extra information is known there are multiple possibilities. A right triangle with all sides known but angles missing Use any inverse trig function Non-Right Triangles use Law of Sines or Law of Cosines Use Law of Sines with 2 angles + 2 sides with one part missing. Often must first use sum of angles is 180°. c A B Law of Sines a b Use Law of Cosines with 3 sides known and missing angles or side-angle-side combo known with side opposite angle missing Law of Cosines sin a sin b sin c = = A B C C c A B a b A2 + B2 – 2ABcos θ = C2 C cos θ = A2 + B2 − C2 2AB Examples 3 Sides known, angles Missing 2 Side-Opposite Angle Combos with 3 parts known Use Law of Cosines to find first angle Many choices after that side-angle-side combo known Use Law of Sines to find missing part Many choices after that Use Law of Cosines to find third side angle-angle-side combo known angle-side-angle combo known rd Use 180° in triangle for 3rd angle Use Law of Sines to find sides Use 180° in triangle for 3 angle Use Law of Sines to find sides Be careful when using Law of Sines with Obtuse Triangles! Summary of Basic Trigonometry: Using Trigonometry in a geometry problem involves triangles. The triangle(s) may be obvious but sometimes they are imbedded deceptively within a more complicated geometry. Generally, the task at hand is to use partial information given about the various triangles to determine all three angles (A) and all three sides (S) of each triangle. A S S A A S Given only SSS, SAS, ASA, AAS we always have a unique triangle. Thus, the remaining three parts of the triangle are predetermined. In each of the examples below the remaining three parts (?) can be found using trigonometry formulas. ? S A ? S ? S ? S SAS ? S SSS ? ? ? ? A S ASA A A ? A ? S ? AAS However, if given SSA two possible triangles can occur but only two. Trigonometry can still give us the unknown parts exactly if we know which of the two cases we have, an acute triangle or an obtuse triangle. S S A S S A 2 possibilities acute triangle obtuse triangle If we have just the three angles known there are an infinite number of similar triangles. Thus, we must always have at least one side given to exactly determine all parts. A ? ? A With the exception of AAA (similar triangles) knowing three of the six triangle parts allows us to find the remaining three parts (assuming we have a sketch for the SSA case). ? A In the case of right triangles, one angle (the right angle) is already known so we only need two additional parts to exactly determine all six parts. Right triangles are usually preferable because the relationships are simpler to work with. Not every triangle requires trigonometry to determine its components. Often a triangle is oriented so that "slope" makes sense for one "angle" of the triangle. If a slope is given (or desired) we can often avoid trigonometry and still determine all the remaining parts of the triangle. In Trigonometry Applications: 1) Find triangles inherent and useful to the problem. Label them. 2) Decide if a triangle is a right-triangle or non-right-triangle. 3) Determine which relation is required. 4) Use the relationship to set up an equation. 5) Solve the equation 6) Repeat as necessary Some hints: Only right triangles can use the Pythagorean Theorem. Recall there are 180° in a triangle and complementary angles sum to 90°. Use Sin, Cos or Tan with right triangles when the angle is known. Use Sin–1, Cos–1 or Tan–1 with right triangles when the angle is the unknown. When two sides form a "t" use Tan or Tan–1. Right-triangles which involve the hypotenuse and an angle use either Sin or Cos. Right-triangles which involve two sides bracketing an angle use Cos. Right-triangles which involve the hypotenuse and a side opposite an angle use Sin. Non-right triangles use Law of Sines or Law of Cosines. When a right triangle is in its standard position the various relationships are easier to recognize and apply. standard position with slope standard position with angle a c cm m= ; a2 + b2 = c2; b= a= 2; b 1+m 1 + m2 tan θ = rise ; run sin θ = rise ; hyp a = c sin θ cos θ = run hyp b = c cos θ When a right triangle is in a random position be careful when using the trigonometric relationships. be careful using slope here a m = tan θ = b opp a = tan–1 adj b sin θ = opp b = hyp c opp adj ; θ = cos–1 hyp hyp cos θ = adj a = hyp c θ = tan–1 θ = sin–1 slope is inapplicable here opp b tan θ = = adj a Procedure Summary: The triangle is a Right Triangle The triangle is NOT a Right Triangle If two sides are known, find the third side by using the Pythagorean Theorem. If three parts of three sides and one angle are known use Law of Cosines to find the fourth part. If one complementary angle is known find the other using complements sum to 90°. A2 + B2 – 2AB cos () = C2 If two sides are known find an angle with: opp adj cos () = A2 + B2 – C2 2AB opp = sin–1 hyp , = cos–1 hyp , = tan–1 adj . If one angle and one side are known find another side using: opp adj opp sin() = hyp , cos() = hyp , tan () = adj If three parts of two sides and two angles are known use Law of Sines to find the fourth part. sin a sin b sin c A = B = C Examples Imbedded Triangles with P||Q Reflected Triangles with P||Q Use Similar Triangles Use Similar Triangles Two Angles Match. Use 180° in triangles Use Similar Triangles Mitered Right Triangles Use Similar Triangles Use Pythagorean Thm. Reference Triangle 3 Sides, Angles Missing Use Law of Cosines to find angles 2 Side-Angle Combos Use Law of Sines to find a Many choices after that Side-Angle-Side Use Law of Cosines to find third side Angle- Angle-Side Use 180° in triangle Use Law of Sines to find sides Angle-Side-Angle Use 180° in triangle Use Law of Sines to find sides Be careful when using Law of Sines with Obtuse Triangles!