Trigonometry & Inverse Trigonometry
Trigonometry
star_batch_jee_advanced_2025
Grade 11
Question:
Points $D, E$ are taken on the side $BC$ of $\triangle ABC$, such that $BD = DE = EC$ and let $\angle BAD = x, \angle DAE = y, \angle EAC = z$; then $\frac{\sin(x+y)\sin(y+z)}{\sin x \sin z} =$
$4$
$6$
$8$
None of these
Step-by-Step Solution
Key Concept: Apply the sine rule systematically to overlapping triangles formed by cevians, then multiply the ratios to establish the product relationship.
Apply the sine rule to triangles $ADC$, $ABD$, $AEC$, and $ABE$ to express segments in terms of sines of angles. From $\triangle ADC$: $\frac{\sin(y+z)}{DC} = \frac{\sin C}{AD}$; from $\triangle ABD$: $\frac{\sin x}{BD} = \frac{\sin B}{AD}$; from $\triangle AEC$: $\frac{\sin z}{EC} = \frac{\sin C}{AE}$; from $\triangle ABE$: $\frac{\sin(x+y)}{BE} = \frac{\sin B}{AE}$. Multiplying these four equations and simplifying yields $\frac{\sin(x+y)\sin(y+z)}{\sin x \sin z} \cdot \frac{BE \cdot DC \cdot AD \cdot AE}{AE \cdot AD \cdot BD \cdot EC} = 4$.
Correct Answer: 1