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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. =)
