Question

ifsinA = 4/5 find the value of (4+tanA)(2+cosA)​

Answer 1

Answer:

The value of (4+tanA)(2+cosA):

208/15 or 13.866

Step by step explained in photograph

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Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

GIVEN :-

• $\tt sin A =\dfrac{4}{5}$

FIND THE VALUE OF :-

• $\tt(4 + tan A)(2 + cos A)$

SOLUTION :-

$\tt sin A = \dfrac {Perpendicular}{Hypotenuse}$

Given, $\tt {sin A = \dfrac{4}{5}}$

Therefore, Perpendicular of the given right angles triangle is 4k and the Hypotenuse is 5k.

Finding the 3rd side by Pythagoras theorem,

$\leadsto \tt {(5k)}^{2} = {(4k)}^{2} + {(x)}^{2}$

$\leadsto \tt {(x) }^{2} = 9{k}^{2}$

$\leadsto \tt x = 3k$

The third side = 3k.

Finding cos A,

$\longrightarrow {\tt cos A = \dfrac{Base}{Hypotenuse}}$

Putting in the values,

$\longrightarrow\huge{\boxed {\tt cos A = \dfrac {3}{5}}}$

Finding the value of tan A,

$\tt tan A = \dfrac{sin A}{cos A}$

Therefore, $\longrightarrow\tt tan A = \dfrac {\dfrac {4}{5}}{\dfrac {3}{5}}$

$\longrightarrow\tt tan A = \dfrac {4}{\cancel {5}}\times\dfrac {\cancel {5}}{3}$

$\longrightarrow\huge{\boxed {\tt tan A = \dfrac {4}{3}}}$

Putting the values of cos A and tan A in the equation whose value we need to find :-

$\implies \tt (4 + tan A)(2 + cos A)$

$\implies \tt (4 + \dfrac {4}{3})(2 + \dfrac {3}{5})$

$\implies \tt (\dfrac {12 + 4}{3})(\dfrac {10 + 3}{5})$

$\implies \tt (\dfrac {16}{3})(\dfrac {13}{5})$

$\implies \tt \dfrac{208}{15}$

So, the final value is :- $\huge{\boxed{\tt {\dfrac {208}{15}}}}$