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If cot B = $\frac{12}{5}$, what is the value of sec B ? |
$\frac{5}{12}$ $\frac{13}{12}$ $\frac{13}{5}$ $\frac{12}{13}$ |
$\frac{13}{12}$ |
cotB = \(\frac{12}{5}\) Using pythagoras theorem , P2 + B2 = H2 52 + 122 = H2 H2 = 169 H = 13 Now , secB = \(\frac{13}{12}\) |
