In $△ A B C$ ,right angled at B, $A B + A C = 9 c m a n d B C = 3 c m$
The value of $sec C$ is

\frac{4}{3}

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\frac{5}{3}

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\frac{1}{3}

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none

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Solution

The correct option is B $\frac{5}{3}$
Let AB $= x$ , then AC $= 9 - x$
${A C}^{2} = {A B}^{2} + {B C}^{2} {(9 - x)}^{2} = x^{2} + 9 x = 4 c m sec C = \frac{A C}{B C} = \frac{4}{3}$

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