Question

(Trigonometry) If Tan A = 24/7 find Cos A

Answer 1

We know, Tan A = P/B
Here, Perpendicular = 24
Base = 7

We need to calculate Hypotenuse by Pythagoras Theorem, 
H² = 24² + 7²
H² = 576 + 49
H = √625
H = 25

Now, we know, Cos A = B/H = 7/25

In short, Your Answer would be 7/25

Hope this helps!

Answer 2

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$\mathsf{tan\,A=\dfrac{24}{7}}\\\\\\ \mathsf{\dfrac{sin\,A}{cos\,A}=\dfrac{24}{7}}\\\\\\ \mathsf{7\,sin\,A=24\,cos\,A}$


Square both sides:

$\mathsf{(7\,sin\,A)^2=(24\,cos\,A)^2}\\\\ \mathsf{49\,sin^2\,A=576\,cos^2\,A}\qquad\quad\textsf{(but }\mathsf{sin^2\,A=1-cos^2\,A}\textsf{)}\\\\ \mathsf{49\cdot (1-cos^2\,A)=576\,cos^2\,A}\\\\ \mathsf{49-49\,cos^2\,A=576\,cos^2\,A}\\\\ \mathsf{49=576\,cos^2\,A+49\,cos^2\,A}$

$\mathsf{49=625\,cos^2\,A}\\\\ \mathsf{cos^2\,A=\dfrac{49}{625}}\\\\\\ \mathsf{cos\,A=\pm\sqrt{\dfrac{49}{625}}}\\\\\\ \mathsf{cos\,A=\pm\sqrt{\dfrac{7^2}{25^2}}}\\\\\\ \mathsf{cos\,A=\pm\,\dfrac{7}{25}}$


Since $\mathsf{tan\,A=\dfrac{24}{7},}$ which is positive, then $\mathsf{A}$ lies either in the 1st or the 3rd quadrant.


•  If $\mathsf{A}$ is a 1st quadrant arc, then

$\mathsf{cos\,A\ \textgreater \ 0\quad\Rightarrow\quad cos\,A=\dfrac{7}{25}\qquad\quad\checkmark}$


•  If $\mathsf{A}$ is a 3rd quadrant arc, then

$\mathsf{cos\,A\ \textless \ 0\quad\Rightarrow\quad cos\,A=-\,\dfrac{7}{25}\qquad\quad\checkmark}$


I hope this helps. =)