Question

In triangle ABC, right angled at B, If tan A = 4/3 , then the value of cosC is​

Answer 1

Answer:

The answer is $\frac{4}{5}$

Step-by-step explanation:

Since B is right angle

$\Rightarrow A+C=90^\circ$

$\Rightarrow C=90^\circ -A$

Also since $\tan A=\frac{4}{3}$

$\Rightarrow \sec^2A=1+\tan^2A=1+\frac{16}{9}=\frac{25}{9}\Rightarrow \sec A=\frac{5}{3}\\\Rightarrow \cos A=\frac{1}{\sec A}=\frac{3}{5}\\\Rightarrow \sin^2A=1-\cos^2A=1-\frac{9}{25}=\frac{16}{25}\\\Rightarrow \sin A=\frac{4}{5}$

$\therefore \cos C=\cos(90^\circ-A)=\sin A=\frac{4}{5}$

(This could be easily done by finding hypotenuse also)

Answer 2

Answer:

Step-by-step explanation:

In triangle ABC, right angled at B, If tan A = 4/3 , then the value of cosCis 3 by 5