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ΔPQR is a right angled triangle. ∠Q = 90 degree, PQ = 9 cm and QR = 11 cm. What is the value of cot P ? |
$\frac{9}{11}$ $\frac{11}{9}$ $\frac{9}{\sqrt{202}}$ $\frac{\sqrt{202}}{9}$ |
$\frac{9}{11}$ |
ΔPQR is right-angled at Q PQ = 9 cm and QR = 11 cm So, cot P = \(\frac{Base}{Perpendicular}\) = \(\frac{9}{11}\) |
