Question

Find the value of sec 45degree geomentrically

Answer 1

     

$\sf{Draw \triangle ABC \ \sf{with} \ \angle A = \angle B = 45 \textdegree\ \sf{and} \ \angle C=90 \textdegree. \\ \\ \therefore\ AC = BC = 1 \ \sf{unit} \ \\ \\ \\ \sf{Here,} \ \ AB=\sqrt{AC^2+BC^2}=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2}$

$\sec A=\frac{\sf{hypotenuse}}{\sf{adjacent\ side}} \\ \\ \\ \sec A = \frac{AB}{AC} \\ \\ \\ \sec 45=\frac{\sqrt{2}}{1} \\ \\ \\ \sec 45=\sqrt{2}$

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