7no. plzzz can u do it
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if 1-cosA,=1/2 =)cos60=1/2
so sin60=√3/2 and tan 60=√3
hence tan 2a+4sin2a=3+3=6
so sin60=√3/2 and tan 60=√3
hence tan 2a+4sin2a=3+3=6
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Cos a = 1/2
B/H =1/2
Let base be x and hypotnues be 2x
Perpendicular = under root of 2x ka whole square-x ka whole square
Perpendicular =x root 13
Tan square A+4sin square A =6
(P/B)2+ 4(P/H) 2=6
(X root 3 / x) 2+4 (x root 3/2x)2=6
(Root 3)2+4(root 3/2)2=6
3+3=6
6+6
B/H =1/2
Let base be x and hypotnues be 2x
Perpendicular = under root of 2x ka whole square-x ka whole square
Perpendicular =x root 13
Tan square A+4sin square A =6
(P/B)2+ 4(P/H) 2=6
(X root 3 / x) 2+4 (x root 3/2x)2=6
(Root 3)2+4(root 3/2)2=6
3+3=6
6+6
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