Question

Evaluate \( \sin 210^\circ \cdot \cos 240^\circ \cdot \tan 150^\circ \):

• A. \( \dfrac{-1}{4\sqrt{3}} \)
• B. \( \dfrac{1}{2\sqrt{3}} \)
• C. \( \dfrac{-1}{2\sqrt{3}} \)
• D. \( \dfrac{1}{4\sqrt{3}} \)

Hint:

Remember the signs of trigonometric functions in different quadrants: - In Quadrant I: all positive. - In Quadrant II: sine is positive, others negative. - In Quadrant III: tangent is positive, others negative. - In Quadrant IV: cosine is positive, others negative.

Answer 1

The Correct Option is A

We need to evaluate \( \sin 210^\circ \cdot \cos 240^\circ \cdot \tan 150^\circ \). Step 1:
\( \sin 210^\circ = -\dfrac{1}{2} \) because 210° lies in the third quadrant.

Step 2:
\( \cos 240^\circ = -\dfrac{1}{2} \) because 240° lies in the third quadrant.

Step 3:
\( \tan 150^\circ = -\dfrac{1}{\sqrt{3}} \) because 150° lies in the second quadrant. \[ \sin 210^\circ \cdot \cos 240^\circ \cdot \tan 150^\circ = \left(-\dfrac{1}{2}\right) \cdot \left(-\dfrac{1}{2}\right) \cdot \left(-\dfrac{1}{\sqrt{3}}\right) = \dfrac{-1}{4\sqrt{3}} \]