Question

In a right triangle ABC,if tan A=4/3,then find the value of sec A​

Answer 1

FINAL ANSWER:-

$secA=\frac{5}{3}$

GIVEN:-

right triangle ABC $tanA=\frac{4}{3}$

TO FIND:-

value of sec A

FORMULAS USED:-

⇒ $tanA=\frac{perpendicular}{base}$

⇒ $secA=\frac{hypotenuse}{base}$

$(hypogenous)^{2} =(perpendicular) ^{2}+(base)^{2}$

SOLUTION:-

$tanA=\frac{4}{3}$ ⇒  $tanA=\frac{BC}{AB}$ ⇒ $tanA=\frac{perpendicular}{base}$

$secA=\frac{AC}{AB}$ ⇒ $secA=\frac{hypotenuse}{base}$ ⇒ $secA=\frac{AC}{3}$

as here perpendicular is 4 and base is 3 , but hypotenuse (CA) is ?

so using pythagoras theorem find AC

$AC^{2}=BC^{2}+AB^{2}\\AC^{2}=4^{2}+3^{2}\\AC^{2}=16+9\\AC^{2}=25\\AC=\sqrt{25}\\AC=5$

as we get AC lets put in  $secA=\frac{AC}{3}$

$secA=\frac{5}{3}$

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Answer 2

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