JEE Mains · Maths · STD 11 - Trigonometrical equations
\(ABCD \) is a trapezium such that \(AB\) and \(CD \) are parallel and \(BC\; \bot CD\). If \(\angle ADB = \theta \),\(BC=p\) and \( CD=q\) , then \(AB\) is equal to:
- A \(\frac{{\left( {{p^2} + {q^2}} \right)\sin \theta }}{{pcos\;\theta + qsin\theta }}\)
- B \(\;\frac{{\;{p^2} + {q^2}\cos \theta }}{{pcos\;\theta + qsin\theta }}\)
- C \(\;\frac{{\left( {{p^2} + {q^2}} \right)}}{{{p^2}cos\;\theta + {q^2}sin\theta }}\)
- D \(\;\frac{{\left( {{p^2} + {q^2}} \right)\sin \theta }}{{{{\left( {pcos\;\theta + qsin\theta } \right)}^2}}}\)
Answer & Solution
Correct Answer
(A) \(\frac{{\left( {{p^2} + {q^2}} \right)\sin \theta }}{{pcos\;\theta + qsin\theta }}\)
Step-by-step Solution
Detailed explanation
Let \(AB=x\) \(\tan \,(\pi \, - \,\theta \, - \,\alpha )\, = \,\frac{p}{{x - q}}\, \Rightarrow \,\tan \,(\theta \, + \,\alpha )\, = \,\frac{p}{{q - x}}\) \( \Rightarrow \,q\, - \,x\, = p\,\,\cot \,(\theta \, + \,\alpha )\)…
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