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प्रश्न
If seс (90° − 3A) . cos 48° = 1 and 0 ≤ 3A ≤ 90°; find the value of angle A.
बेरीज
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उत्तर
Given:
seс (90° − 3A) . cos 48° = 1 and 0 ≤ 3A ≤ 90°
Formula:
\[\sec(\theta) = \frac{1}{\cos(\theta)} \text{ and } \sec(90^\circ - \theta) = \csc(\theta)\]
Solution:
\[\sec(90^\circ - 3A) = \frac{1}{\cos 48^\circ}\] [Transposing cos 48° to the right side]
seс (90° − 3A) = sec 48° [Using the reciprocal identity \[\frac{1}{\cos 48^\circ} = \sec 48^\circ\]]
90° − 3A = 48° [Equating the angles of equal secant functions]
3A = 90° − 48° [Rearranging terms to solve for 3A]
3A = 42° [Subtracting 48° from 90°]
\[A = \frac{42^\circ}{3}\] [Dividing both sides by 3]
A = 14° [Simplifying the division]
∴ A = 14°
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