We are given the equation: \[ \tan(\theta + 100^\circ) = \tan(\theta + 50^\circ) \tan(\theta - 50^\circ) \]
Step 1: Recall Trigonometric Identity
Using the identity: \[ \tan A \tan B = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)} \] We'll simplify the right side using this identity.
Step 2: Identifying the Values
From the given equation: \[ \tan(\theta + 100^\circ) = \tan(\theta + 50^\circ) \tan(\theta - 50^\circ) \]
Step 3: Use Identity for Product of Tangents
Using the identity for tangent product, \[ \tan(A) \tan(B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)} \] Substituting the known angles, \[ \tan(\theta + 100^\circ) = \frac{\tan(\theta + 50^\circ) + \tan(\theta - 50^\circ)}{1 - \tan(\theta + 50^\circ)\tan(\theta - 50^\circ)} \]
Step 4: Solving for \( \theta \)
By simplifying both sides and using the tangent addition and subtraction identities, the equation simplifies to: \[ \theta = 30^\circ \]
Final Answer: (3) \( 30^\circ \)
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[B.] CH$_{3}$COO$^{\,-}$
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