if cosA = 2/5 find 4 +4tansquareA
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Answered by
2
cosA = 2/5 = b/h
h^2 = b^2+p^2
5^2 = 2^2+p^2
25 = 4 +p^2
p^2 = 21
p = root of 21
= 4+4tan^2A
= 4 + 4*21/4
= 4+21
= 25
h^2 = b^2+p^2
5^2 = 2^2+p^2
25 = 4 +p^2
p^2 = 21
p = root of 21
= 4+4tan^2A
= 4 + 4*21/4
= 4+21
= 25
Answered by
1
Cos A=base/hypotenuse =2/5
Then hypotenuse =5x, base=2x
By Pythagoras theorem
25x²-4x²=hieght²
21x²=hieght²
√21x²=hieght
Tan²A=hieght²/base² =21x²/4x²=21/4
4+4*21/4 =4+21=25
25 is your answer
Then hypotenuse =5x, base=2x
By Pythagoras theorem
25x²-4x²=hieght²
21x²=hieght²
√21x²=hieght
Tan²A=hieght²/base² =21x²/4x²=21/4
4+4*21/4 =4+21=25
25 is your answer
InnocentBachiNo1:
but correct answer is 25
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