Question

proof that (i)sin45°=1/√2 in
geometry way
(ii)Sin(90°-A)=COS A


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Answer 1

Step-by-step explanation:

this is the best answer

Answer 2

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Step-by-step explanation:

In a triangle ABC

angle B = 90

If one angle is 45 then the other angle will also be 45

i.e angle A = angle C = 45

=> BC = AB

by Pythagoras theorem

$h^{2} = p^{2} + b^{2}\\\\h^{2} = a^{2} + a^{2}\\\\h^{2} = 2a^{2}\\\\h= \sqrt{2a^{2}}\\\\h= a\sqrt{2}$

sin 45 = $\frac{p}{h}$

$= \frac{BC}{AC} \\\\= \frac{a}{a\sqrt{2} } \\\\= \frac{1}{\sqrt{2} }$

ii)  In triangle ABC,

sin A = $\frac{p}{h}$

$= \frac{BC}{AC}$

cos A = $\frac{b}{h}$

$= \frac{AB}{AC}$                   --------- (equation 1)

Now,

angle A + angle C = 90

angle C = 90 - angle A

=> sin (90 - A) = angle C

If theta is at C then

p = AB

&

h = AC

Therefore

sin (90-A) = $\frac{AB}{AC}$                      -----  (equation 2)

Now compare equation 1 and 2

We conclude that

sin(90 - A) = cos A