Question

CLASS 10
MATHS
TRIGNOMETRY...

1. Find the value to sin30° geometrically.


2. Find the value of 60° geometrically.


Answer this question by solving in copy....


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Answer 1

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Answer 2

$\huge\bold{ANSWER-1--}$

For finding the value of sin 30° geometrically, we construct an equilateral triangle ABC  of side 2 a. From A, we draw perpendicular AD to BC. Now, BD = CD  = 1/2 BC = a

In triangle ABD, BAD + BDA + DAB  = 180 (BY ANGLE SUM PROPERTY)

=> 60°+ BAD + 90°= 180°   (SINCE AD | BC  AND ALL ANGLES ARE 60° in a equilateral triangle)

=> BAD = 30°

IN RT TRIANGLE BAD,

SIN 30° =  BD / AB=  a / 2 a   = 1/ 2

hence geometrically shown that sin 30 is 1/2

$\huge\bold{ANSWER-2--}$

ABC in which perpendicular to AD .

BD = a and DC = a and AB = 2a.

Now in ∆ABD

AB^2 = Ad^2 + BD ^ 2

( 2a )^ 2 = AD ^ 2+ a ^2

4a^2 - a ^2 = AD ^2

3a ^2 = AD^2

√3a = AD

Then ,
sin 60= p/h

= √3a/ 2a

= √3/ 2

since , Sin60° = √3/2