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If sin A = $\frac{6}{7}$, then what is the value of sec A ? |
$\frac{1}{\sqrt{13}}$ $\frac{7}{\sqrt{13}}$ $\frac{6}{\sqrt{13}}$ $\frac{4}{\sqrt{13}}$ |
$\frac{7}{\sqrt{13}}$ |
sinA = \(\frac{6}{7}\) { we know that sinθ = \(\frac{P}{H}\) } By using pythagoras theorem , P² + B² = H² 6² + B² = 7² B² = 49 - 36 B² = 13 B = \(\sqrt {13 }\) Now, secA = \(\frac{H}{B}\) = \(\frac{7}{√13}\) |
