If sinA=7/10 find the value of 5 5cot^2a
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Answered by
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Sin A = perpendicular / hypotenius
7/10 = p / h
by making the triangle
and on applying pythagorus
h^2 = p^3 + b^2
100 = 49 + b^2
100-49 = b^2
b = root 51
Cot a = b / p
cot a = root 51 / 7
on putting in the given question
=5 x 5 (root 51 / 7)^2
=5 x 5 (51 / 49)
=25 x 51 / 49
=1275 / 49
=26.02
7/10 = p / h
by making the triangle
and on applying pythagorus
h^2 = p^3 + b^2
100 = 49 + b^2
100-49 = b^2
b = root 51
Cot a = b / p
cot a = root 51 / 7
on putting in the given question
=5 x 5 (root 51 / 7)^2
=5 x 5 (51 / 49)
=25 x 51 / 49
=1275 / 49
=26.02
Answered by
0
Given :
sinA=7/10
so, cosecA=10/7
To find :
5+5cot²A
Solution:
=5+5cot²A
=5(1+cot²A)
[Using cosec²A=1+cot²A]
=5cosec²A
=5×(10/7)²
=5×(100/49)
=500/49
sinA=7/10
so, cosecA=10/7
To find :
5+5cot²A
Solution:
=5+5cot²A
=5(1+cot²A)
[Using cosec²A=1+cot²A]
=5cosec²A
=5×(10/7)²
=5×(100/49)
=500/49
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