# Trigonometry Exit Definitions Further topics Further topics sine cosine & tangent Triangle terminology Triangle terminology adjacent side opposite side.

## Presentation on theme: "Trigonometry Exit Definitions Further topics Further topics sine cosine & tangent Triangle terminology Triangle terminology adjacent side opposite side."— Presentation transcript:

Trigonometry Exit Definitions Further topics Further topics sine cosine & tangent Triangle terminology Triangle terminology adjacent side opposite side angle hypotenuse Pythagoras’ theorem Pythagoras’ theorem a 2 + b 2 = c 2 sin 2 (  ) + cos 2 (  ) = 1

Triangle terminology Exit Opposite side Opposite side Adjacent side Adjacent side Hypotenuse Contents  terminology  terminology Definitions

angle Opposite side: the side opposite the angle Exit Clear opposite Hypotenuse Adjacent side Adjacent side opposite Contents  terminology  terminology Definitions angle

adjacent Adjacent side: the side beside the angle Exit Opposite side Opposite side Clear Hypotenuse adjacent Contents  terminology  terminology Definitions angle

Hypotenuse: the longest side Exit Opposite side Opposite side Clear Adjacent side Adjacent side hypotenuse Contents  terminology  terminology Definitions

Exit Tangent Sine Cosine Contents  terminology  terminology A B 90 o C a c b Definitions

Exit clear Sine Cosine Contents  terminology  terminology A B 90 o C a c b Definitions tan(A) = tan(B) Tangent of angle A opposite adjacent abab =

Tangent of angle B Exit clear Sine Cosine Contents  terminology  terminology A B 90 o C a c b tan(B) = Definitions tan(A) opposite adjacent baba =

Sine of angle A Exit Tangent Cosine Contents  terminology  terminology A B 90 o C a c b Definitions sin(A) == acac sin(B) opposite hypotenuse Clear

Sine of angle B Exit Tangent Clear Contents  terminology  terminology A B 90 o C a c b Definitions sin(B) == bcbc sin(A) opposite hypotenuse Cosine

Cosine of angle A Exit Tangent Sine Clear Contents  terminology  terminology A B 90 o C a c b Definitions cos(A) == bcbc cos(B) adjacent hypotenuse

Cosine of angle B Exit Tangent Sine Clear Contents  terminology  terminology A B 90 o C a c b Definitions cos(B) == acac cos(A) adjacent hypotenuse

Trigonometry: further topics Exit sin(A+B) = sin(A)cos(B) + cos(A)sin(B) Identity Cosine rule Cosine rule Sine rule Sine rule Contents Further topics Further topics a sin(A) b sin(B) = c sin(C) = cos(A) = b 2 + c 2 - a 2 2bc Area = ½ a b sin(C)= ½ b c sin(A) = ½ c a sin(B) Area of triangle Area of triangle

Sine Rule: Exit Contents AC Further topics Further topics ca b A & B A & B & C B & C back Clear B sin(A) a = sin(C) c sin(B) b =

Sine Rule: Exit Contents Further topics Further topics back Clear C ca b A h B sin(A) ==== a = sin(C) c sin(B) b =

Sine Rule: Exit Contents A Further topics Further topics ca b back Clear h B sin(A) h b = a => ==== sin(A) a = sin(C) c sin(B) b = C

Sine Rule: Exit Contents Further topics Further topics back Clear A ca b h B ==== sin(A) a = sin(B) sin(A) hbhb => h b a = sin(A) a = sin(C) c sin(B) b = C

Sine Rule: Exit Contents Further topics Further topics back Clear ==== A ca b h B sin(A) a = sin(B) b sin(A) hbhb h a => h b a == sin(A) a = sin(C) c sin(B) b = C

Sine Rule: Exit Contents sin(A) a = Further topics Further topics sin(B) b sin(B)sin(A) hbhb haha => back Clear => A a b h B h b a h a b = == => sin(A) a = sin(C) c sin(B) b = C c

Sine Rule: Exit Contents sin(A) a = Further topics Further topics sin(B) b sin(B)sin(A) hbhb haha => back Clear => A a b h B sin(A) a = sin(C) c sin(B) b = h b a h a b = == sin(B) b sin(A) a => = C c

Sine Rule: Exit Contents Further topics Further topics back Clear ca b B sin(B) ==== sin(A) a = sin(C) c sin(B) b = C A h

Sine Rule: Exit Contents Further topics Further topics back Clear sin(B) h c = b => ==== sin(A) a = sin(C) c sin(B) b = ca b B C A h

Sine Rule: Exit Contents Further topics Further topics back Clear ==== sin(B) b = sin(C) sin(B) hchc => h c b = sin(A) a = sin(C) c sin(B) b = ca b B C A h

Sine Rule: Exit Contents Further topics Further topics back Clear ==== sin(B) b = sin(C) c sin(B) hchc h b => h c b == sin(A) a = sin(C) c sin(B) b = ca b B C A h

Sine Rule: Exit Contents sin(B) b = Further topics Further topics sin(C) c sin(C)sin(B) hchc hbhb => back Clear => h c b h b c = == => sin(A) a = sin(C) c sin(B) b = ca b B C A h

Sine Rule: Exit Contents sin(B) b = Further topics Further topics sin(C) c sin(C)sin(B) hchc hbhb => back Clear => sin(A) a = sin(C) c sin(B) b = h c b h b c = == sin(C) c sin(B) b => = ca b B C A h

sin(P+Q) : Exit Contents Further topics Further topics back Clear P Q = sin(P) cos(Q) + sin(Q) cos(P) Show proof Show proof

sin(P+Q) : Exit Contents Further topics Further topics => sin(P+Q) = back = sin(P) cos(Q) + sin(Q) cos(P) p b B P q Q Sine rule Sine rule Clear

sin(P+Q) : Exit Contents Further topics Further topics => sin(P+Q) = (p+q) back sin(B) b = sin(P) cos(Q) + sin(Q) cos(P) => sin(P+Q) = Substitute sin(B) = cos(P) Substitute sin(B) = cos(P) p b q c h ;sin(B)== cos(P) hchc B P Q Clear

sin(P+Q) : Exit Contents Further topics Further topics => sin(P+Q) =(p+q) back sin(B) b = sin(P) cos(Q) + sin(Q) cos(P) p b B P q Q => sin(P+Q) =(p+q) cos(P) b c h => sin(P+Q) = Expand bracket Expand bracket ;sin(B)== cos(P) hchc Clear

sin(P+Q) : Exit Contents Further topics Further topics => sin(P+Q) =(p+q) back sin(B) b = sin(P) cos(Q) + sin(Q) cos(P) b => sin(P+Q) =(p+q) cos(P) b => sin(P+Q) = pbpb cos(P) qbqb + => sin(P+Q) = Substitute: cos(P) = h / c ; q / b =sin(Q) ;cos(P)= hchc =sin(Q) qbqb p b B P q Q c h Clear

sin(P+Q) : Exit Contents Further topics Further topics => sin(P+Q) =(p+q) back sin(B) b = sin(P) cos(Q) + sin(Q) cos(P) => sin(P+Q) =(p+q) cos(P) b => sin(P+Q) = pbpb cos(P) qbqb + => sin(P+Q) = pbpb sin(Q) + hchc x => sin(P+Q) = cos(P) Substitute: p / c = sin(P) ; h / b = cos(Q) p b B P q Q c h = sin(P) pcpc = cos(Q) hbhb ;, Clear

sin(P+Q) : Exit Contents Further topics Further topics => sin(P+Q) =(p+q) back sin(B) b = sin(P) cos(Q) + sin(Q) cos(P) p b B P q Q => sin(P+Q) =(p+q) cos(P) b c h => sin(P+Q) = pbpb cos(P) qbqb + => sin(P+Q) = pbpb sin(Q)+ hchc x => sin(P+Q) = cos(P) sin(Q)cos(P)+ sin(P)cos(Q) = sin(P) pcpc = cos(Q) hbhb ;, Clear

Pythagoras’ theorem: Exit Contents Further topics Further topics back Clear sin 2 (A) +cos 2 (A) = 1 sin(A) A B Cb a c ==== a 2 + b 2 = c 2

A Pythagoras’ theorem: Exit Contents Further topics Further topics back sin(A) B Cb a c x y acac = ==== A Clear sin 2 (A) +cos 2 (A) = 1 a 2 + b 2 = c 2

Pythagoras’ theorem: Exit Contents Further topics Further topics back sin(A) B Cb a c x y acac y a = =>a 2 = cy = cos(A) ==== A A Clear sin 2 (A) +cos 2 (A) = 1 a 2 + b 2 = c 2

==== Pythagoras’ theorem: Exit Contents Further topics Further topics back sin(A) B Cb a c x y acac yaya ==>a 2 = cy= cos(A) bcbc = A Clear sin 2 (A) +cos 2 (A) = 1 a 2 + b 2 = c 2

Pythagoras’ theorem: Exit Contents Further topics Further topics back sin(A) B Cb a c x y acac yaya ==>a 2 = cy= cos(A) bcbc xbxb ==>b 2 = cx= => A Clear sin 2 (A) +cos 2 (A) = 1 a 2 + b 2 = c 2

Pythagoras’ theorem: Exit Contents Further topics Further topics back sin(A) B Cb a c x y acac yaya ==>a 2 = cy= cos(A) bcbc xbxb ==>b 2 = cx= => a 2 + b 2 = cy +cx => A Note: y+x=c => c(y + x ) = c2c2 Clear sin 2 (A) +cos 2 (A) = 1 a 2 + b 2 = c 2

Pythagoras’ theorem: Exit Contents Further topics Further topics back sin(A) B Cb a c x y acac yaya ==>a 2 = cy= cos(A) bcbc xbxb ==>b 2 = cx= =>a 2 + b 2 = cy +cx a 2 + b 2 = c2c2 => Note: y+x=c => c(y + x ) = c 2 A Clear sin 2 (A) +cos 2 (A) = 1 a 2 + b 2 = c 2

a2a2 Pythagoras’ theorem: Exit Contents Further topics Further topics back sin(A) B Cb a c x y acac yaya ==>a 2 = cy= cos(A) bcbc xbxb ==>b 2 = cx= =>a 2 + b 2 = cy +cx = c(y + x ) = c 2 b2b2 c2c2 +==> c2c2 c2c2 c2c2 a 2 + b 2 = c 2 => A Clear sin 2 (A) +cos 2 (A) = 1 a 2 + b 2 = c 2

a2a2 Pythagoras’ theorem: Exit Contents Further topics Further topics back sin(A) B Cb a c x y acac yaya ==>a 2 = cy= cos(A) bcbc xbxb ==>b 2 = cx= =>a 2 + b 2 = cy +cx = c(y + x ) = c 2 b2b2 c2c2 += + cos 2 (A) => sin 2 (A)= 1 c2c2 c2c2 c2c2 =>a 2 + b 2 = c 2 => A Clear sin 2 (A) +cos 2 (A) = 1 a 2 + b 2 = c 2

Show proof Show proof Cosine Rule: Exit Contents A C Further topics Further topics ca b back Clear B cos(A)= b 2 + c 2 - a 2 2bc a 2 = b 2 + c 2 - 2bc cos(A)

Pythagoras’ Theorem Pythagoras’ Theorem Cosine Rule: Exit Contents A C Further topics Further topics ca b back B cos(A)= b 2 + c 2 - a 2 2bc a2a2 xb-x h a 2 = b 2 + c 2 - 2bc cos(A) Clear =

Expand bracket Expand bracket Cosine Rule: Exit Contents Further topics Further topics back cos(A)= b 2 + c 2 - a 2 2bc a2a2 =(b-x) 2 +h2h2 a2a2 => a 2 = b 2 + c 2 - 2bc cos(A) Clear A C ca b B xb-x h =

Substitute h 2 = c 2 - x 2 Substitute h 2 = c 2 - x 2 Cosine Rule: Exit Contents A C Further topics Further topics ca b back B cos(A)= b 2 + c 2 - a 2 2bc a2a2 xb-x h =(b-x) 2 +h2h2 a2a2 =+h2h2 => a2a2 b 2 - 2bx+ x 2 a 2 = b 2 + c 2 - 2bc cos(A) Note: h 2 = c 2 - x 2 Clear =

Simplify Cosine Rule: Exit Contents A C Further topics Further topics ca b back B cos(A)= b 2 + c 2 - a 2 2bc a2a2 xb-x h =(b-x) 2 +h2h2 a2a2 =+h2h2 => a2a2 =b 2 - 2bx+c2c2 =>- x 2 + x 2 b 2 - 2bx+ x 2 a2a2 => a 2 = b 2 + c 2 - 2bc cos(A) Note: h 2 = c 2 - x 2 Clear =

Substitute x = c cos(A) Substitute x = c cos(A) Cosine Rule: Exit Contents A C Further topics Further topics ca b back B cos(A)= b 2 + c 2 - a 2 2bc a2a2 xb-x h =(b-x) 2 +h2h2 a2a2 =+h2h2 => a2a2 =b 2 - 2bx+c2c2 =>- x 2 + x 2 b 2 - 2bx+ x 2 a2a2 =b 2 - 2bx+c2c2 => a2a2 a 2 = b 2 + c 2 - 2bc cos(A) Note: x = c cos(A) Clear =

Cosine Rule: Exit Contents A C Further topics Further topics ca b back B cos(A)= b 2 + c 2 - a 2 2bc a2a2 xb-x h =(b-x) 2 +h2h2 a2a2 =+h2h2 => a2a2 =b 2 - 2bx+c2c2 =>- x 2 + x 2 b 2 - 2bx+ x 2 a2a2 =b 2 - 2bx+c2c2 => a2a2 = b 2 - 2bc cos(A) +c2c2 => a 2 = b 2 + c 2 - 2bc cos(A) Rearrange Note: x = c cos(A) Clear a2a2 ==>

Cosine Rule: Exit Contents A C Further topics Further topics ca b back B cos(A)= b 2 + c 2 - a 2 2bc a2a2 xb-x h =(b-x) 2 +h2h2 a2a2 =+h2h2 => a2a2 =b 2 - 2bx+c2c2 =>- x 2 + x 2 b 2 - 2bx+ x 2 a2a2 =b 2 - 2bx+c2c2 => a2a2 =b 2 - 2bc cos(A)+c2c2 => a 2 = b 2 + c 2 - 2bc cos(A) a2a2 =b2b2 + c2c2 => - 2bc cos(A) Clear

Area of triangle: Exit Contents A C Further topics Further topics c a b back B Clear Area = ½ a b sin(C)= ½ b c sin(A) = ½ c a sin(B) Next

Area of triangle: Exit Contents A C Further topics Further topics c a b back B height Clear Area = ½ a b sin(C)= ½ b c sin(A) = ½ c a sin(B) Area of rectanglebase x height= Area of triangle ABC= base Next ready

Area of triangle: Exit Contents A C Further topics Further topics c a b back B height Clear Area = ½ a b sin(C)= ½ b c sin(A) = ½ c a sin(B) Area of rectanglebase x height= Area of triangle ABC ½ of LEFT rect. + ½ of RIGHT rect. = base Area of triangle½ base x height= Next ready

Area of triangle: Exit Contents A C Further topics Further topics c a b back B Clear Area = ½ a b sin(C)= ½ b c sin(A) = ½ c a sin(B) Area of triangle½ base x height= height base Next ready

Area of triangle: Exit Contents A C Further topics Further topics c a b back B Clear Area = ½ a b sin(C)= ½ b c sin(A) = ½ c a sin(B) Next Area of triangle½ base x height= height base Area of triangle½ base x=c height c x ready

Area of triangle: Exit Contents A C Further topics Further topics c a b back B Clear Area = ½ a b sin(C)= ½ b c sin(A) = ½ c a sin(B) Area of triangle½ base x height= height base Area of triangle½ base x=c height c x Area of triangle½=cSin(A) b complete

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