# dnl exposé

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Al-Kashi

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Ulugh Beg observatory In Samarkand

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P ythagoras T heorem A²=B²+C² If the triangle had a right angle, and you made a square on each of the three sides , then the biggest square had the exact same area as the other two squares put together .

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a²=b²+c²-2bc×cos(Â) b²=a²+c²-2ac×cos(B) c²=b²+a²-2ba×cos(C)

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How did he do? a²=b²+c²-2bc cos Â D E P I K L M N

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How did he do? a²=b²+c²-2bc cos Â h h E D P I K L M N

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How did he do? a²=b²+c²-2bc cos Â E D P I K L M N

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How did he do? a²=b²+c²-2bc cos Â E D P h I K L M N

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How did he do? a²=b²+c²-2bc cos Â E D P I K L M N

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How did he do? a²=b²+c²-2bc cos Â A3 A1 A2 A1+A2 =A3 A4 A5 AB²=BC²+AC²-(A4+A5)

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How did he do? a²=b²+c²-2bc cos Â BC x E X = AC × cosα A5 A4

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A5 A5 = BC × AC × cosα A5 = BC × X X = AC × cosα

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How did he do? a²=b²+c²-2bc cos Â y E Y = BC × cosα M AC A4 A5

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A4 A4 = AC × BC × cosα A4 = AC × Y Y = BC × cosα

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A4 = AC × BC × cosα A4 = AC × Y Y = CB × cosα A5 = BC × AC × cosα A5 = BC × X X = AC × cosα AB²=BC²+AC²-(A4+A5) AB²=BC²+AC²- AC × BC × cosα - BC × AC × cosα AB²=BC²+AC²- 2AC × BC × cosα

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AB²=BC²+AC²- 2AC × BC × cosα c²=a²+b²- 2ba × cosC

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With the Pythagoras Theorem A B C c b a H x b²=AH²+(a-x)² b²=c²-x²+a²+x²-2ax b²=c²+a²-2ax b²=c²+a²-2ac.cos(ABC)

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With the scalar product A B C c b a We know that : → → AB.AC=AB× AC×cos (BAC) → a²=BC² → → a²=(BA+AC)² → → a²=BA²+AC²+2BA.AC → → a²=c²+b²-2AB.AC ^ a²=b²+c²-2bc×cos(BAC)

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Verification of the Pythagoras Theorem b a c b²=c²+a²-2ac.cos(ABC) And Cos(ABC)=0 So b²=c²+a²

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Exercise A B C c b a I Demonstrate the formula: AI²=c²+a²/4- ac ×cos(B) a²=b²+c²-2bccosÂ b²=a²+c²-2accosB c²=b²+a²-2bacosC

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Exercise A B C c b a I Demonstrate the formula: AI²=a²/4+ bc ×cos(A) a²=b²+c²-2bc×cos(Â) b²=a²+c²-2ac×cos(B) c²=b²+a²-2ba×cos(C) b²+c²=2AI²+a²/2

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Exercise A B C c b a I Express c in terms of a,b,cosB and cosA a²=b²+c²-2bc×cos(Â) b²=a²+c²-2ac×cos(B) c²=b²+a²-2ba×cos(C) b²+c²=2AI²+a²/2 AI²=c²+a²/4- ac ×cos(B) AI²=a²/4+ bc ×cos(A)