find the following figure ABCD triangle the base Y the hypotenuse is to the perpendicular is one
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y is a side of triangle
1 is the other
2 is hypotenuse
thus,
2^2 = y^2 + 1^ 2
4 = y^2 + 1
4= y^2+1
y^2=3
y=√3..(ans(I))
(ii)
sin X= P/H
= y/2
= √3 /2...( ans (ii))
(iii)
(sec X - tanx )(sec X + tan X )
sec^2 X - tan^2 X
H/B^2 - P/B^2
2/1^2 -√3/1^2
4- 3
1..ans (iii)
P - perpendicular
B- Base
H- hypotenuse
( pls. do not forget to follow and thanks)
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here are the answers for (i) (ii) (iii) questions......hope u will understand...
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