If 5sec@ = cosec(@ + 30°) find @
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Sec 5A = cosec (A-30)
A < 90 deg.
we know cosec B = sec(90-B)
so cosec(A - 30) = sec(90 - (A-30))
so sec (120-A) = sec (5A)
so either 120 - A = 5A or 120 - A = - 5A
so A = 20° or -30° or 330°
A < 90 deg.
we know cosec B = sec(90-B)
so cosec(A - 30) = sec(90 - (A-30))
so sec (120-A) = sec (5A)
so either 120 - A = 5A or 120 - A = - 5A
so A = 20° or -30° or 330°
rampalmanohar2011:
it's 5secA not sec5A
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