Question

if cos A is 4/3 than tan A is​

Answer 1

$\sf\huge\underline{Answer :-}$

Given :

• cosA

To find :

• tanA

Solution :

We know that,

$\sf\longrightarrow cosA = \dfrac{Base}{Hypotenuse}$

$\sf\longrightarrow \dfrac{B}{H} = \dfrac{4}{3}$

$\sf\longrightarrow Let \: B \: be \: '4k' \: and \: H \: be \: 3k$

By Pythagoras theorem,

=> H² = P² + B²

=> P² = H² - B²

=> P² = (3k)² - (4k)²

=> P² = 9k² - 16k²

=> P² = -5k²

=> P = √-5k²

=> P = - √5 . k

Now,

$\sf\longrightarrow tanA = \dfrac{Perpendicular}{Base}$

$\sf\longrightarrow tanA = - \dfrac{\sqrt 5.k}{4k}$

$\sf\therefore tanA = - \dfrac{\sqrt 5}{4}$