Solve for 'θ': sec ( θ 2 + 10 ° ) = 2 √ 3

Solve for 'θ': `sec(θ/2 + 10°) = (2)/sqrt(3)`

बेरीज

`sec(θ/2 + 10°) = (2)/sqrt(3)`

⇒ `sec(θ/2 + 10°)` = sec 30°

⇒ `θ/(2) + 10°` = 30°

⇒ `θ/(2)` = 20°
⇒ θ = 40°.

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Trigonometric Equation Problem and Solution

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पाठ 27: Trigonometrical Ratios of Standard Angles - Exercise 27.1

पाठ 27 Trigonometrical Ratios of Standard Angles
Exercise 27.1 | Q 7.3

Solve the following equation for A, if sec 2A = 2

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