If tan 21° 48 = •4. Then 5 sin Φ +2 cos Φ = 5 solved it
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Hay sir, Always good questions
According to the question we know that
tan21°48 = 4 = 2/5
so, sin 21° 48 = 2/√29
cos 21°48 = 5/√29
5 sin Φ +2 cos Φ = 5
If we can divide both sides by √29 then,
5/29 sinΦ + 2/29 cosΦ = 5/29
=> cos21°48 sinΦ + sin21°48 cosΦ = cos21°48
=> sin(Φ+21°48) = cos21°48 = sin 68°12
=> Φ +21°48 = n. 180°+ (-1)^n 68°12
=> Φ = n^180°-21°48 + (-1)^n 68°12
I hope it's help you
Regards : Sangela
According to the question we know that
tan21°48 = 4 = 2/5
so, sin 21° 48 = 2/√29
cos 21°48 = 5/√29
5 sin Φ +2 cos Φ = 5
If we can divide both sides by √29 then,
5/29 sinΦ + 2/29 cosΦ = 5/29
=> cos21°48 sinΦ + sin21°48 cosΦ = cos21°48
=> sin(Φ+21°48) = cos21°48 = sin 68°12
=> Φ +21°48 = n. 180°+ (-1)^n 68°12
=> Φ = n^180°-21°48 + (-1)^n 68°12
I hope it's help you
Regards : Sangela
Shubhendu8898:
great!! but just a point mistake in first Line :-)
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