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If in a triangle tanA plus tanB plus tanC equals 0 then cotAcotBcotC equals?
Kinzer
cotA*cotB*cotC = 1/[tanA+tanB+tanC]
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∙ 10y ago
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How do you solve right spherical triangle when A is equal to 54 degrees 30' B equals 124 degrees 10' and C equals 90 degrees.Find the sides?
Apply Napier's rulesin co-c= (tanco-A)(tanco-B)cosc= 1/tanAtanBc= cos-1(1/tanAtanB)c=118.96 degreessin a= (cosco-A)(cosco-C)a= sin-1(sinAsinC)a=45.52sinb= (tancoA)(tana)b=sin-1 (tana/tanA)b=46.37 or >>>180-46.37 = 133.63FA= c=118.96 degreesa= 45.52 degreesb= 133.63 degrees
If SecASinA equals 0 then what will be the Value of CosA?
secA(sinA)=0 (1/cosA)(sinA)=0 tanA=0 Therefore A is in 1st or 3rd Quadrant i.e A=0 Degrees, 180 Degrees.... This yields cosA=1 or cosA=-1
Why are there two sine cosine and tangent ratios?
There aren't. There are three: Sine, Cosine and Tangent, for any given right-angled triangle. They are related of course: for any given angle A, sinA/cosA = tanA; sinA + cosA =1. As you can prove for yourself, the first by a little algebraic manipulation of the basic ratios for a right-angled triangle, the second by looking up the values for any value such that 0 < A < 90. And those three little division sums are the basis for a huge field of mathematics extending far beyond simple triangles into such fields as harmonic analysis, vectors, electricity & electronics, etc.
If tanA plus tanB plus tanC equals 0 then cotA plus cotB plus cotC equals?
It is (tanA+tanB)/{tanA*tanB} - 1/(tanA+tanB)
How do you solve right spherical triangle when A is equal to 54 degrees 30' B equals 124 degrees 10' and C equals 90 degrees.Find the sides?
Apply Napier's rulesin co-c= (tanco-A)(tanco-B)cosc= 1/tanAtanBc= cos-1(1/tanAtanB)c=118.96 degreessin a= (cosco-A)(cosco-C)a= sin-1(sinAsinC)a=45.52sinb= (tancoA)(tana)b=sin-1 (tana/tanA)b=46.37 or >>>180-46.37 = 133.63FA= c=118.96 degreesa= 45.52 degreesb= 133.63 degrees
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