In triangle ABC, right angled at B, If tan A = 4 3 , then the value of cosC is
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Step-by-step explanation:
The value of CosCCosC is \frac{3}{5}
5
3
Step-by-step explanation:
Given:
A triangle ABC right angled at B
tanA=4/3tanA=4/3
we need to find CosCCosC
In a triangle sum of the angles is 180^0180
0
So we get:
\begin{gathered}A+B+C=180^0\\\\A+C=180^0-90^0\\\\A=90^0-C\end{gathered}
A+B+C=180
0
A+C=180
0
−90
0
A=90
0
−C
Applying tan on both sides
\begin{gathered}Tan A=Tan(90^0-C)\\\\\frac{4}{3}=Cot C\\\\Tan C=\frac{4}{3}\\\end{gathered}
TanA=Tan(90
0
−C)
3
4
=CotC
TanC=
3
4
We know that Sec^2\alpha -Tan^2\alpha =1Sec
2
α−Tan
2
α=1
So we get:
\begin{gathered}Sec^2C=1+(\frac{4}{3} )^{2} \\\\Sec^2C=\frac{25}{9}\\\\SecC=\frac{5}{3} \\\\Cos C=\frac{3}{5}\end{gathered}
Sec
2
C=1+(
3
4
)
2
Sec
2
C=
9
25
SecC=
3
5
CosC=
5
3
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