Question

Find the value of sec45 geometrically. Hence find sin 45 and cos 45 ?

Answer 1

Sin45 = cos45 = 1/squareroot(2)

Answer 2

Answer:

$\sec 45=\sqrt2$

$\sin 45=\frac{1}{\sqrt2}$

$\cos 45=\frac{1}{\sqrt2}$

Step-by-step explanation:

To find : The value of $\sec 45$ geometrically?

Solution :    

Consider a right angle triangle ABC with sides AB=BC=a

$\angle A=\angle C=45^\circ$ and $\angle B=90^\circ$

In triangle ABC, Apply Pythagoras theorem,

$AC^2=AB^2+BC^2$

$AC^2=a^2+a^2$

$AC^2=2a^2$

$AC=a\sqrt2$

So, Perpendicular P=a , Base B=a , Hypotenuse $H=a\sqrt2$

$\sec 45=\frac{H}{P}$

$\sec 45=\frac{a\sqrt2}{a}$

$\sec 45=\sqrt2$

Similarly,

$\sin 45=\frac{P}{H}$

$\sin 45=\frac{a}{a\sqrt2}$

$\sin 45=\frac{1}{\sqrt2}$

$\cos 45=\frac{B}{H}$

$\cos 45=\frac{a}{a\sqrt2}$

$\cos 45=\frac{1}{\sqrt2}$