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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2020 · Mathematics · Trigonometric Ratios & Identities

Paragraph: Read the following information and answer: A B C D is a...

Paragraph: Read the following information and answer: ABCDA B C D is a trapezium such that ABA B and CDC D are parallel and BCB C is perpendicular to them. Let ∠ADB=θ,∠ABD=α,BC=p\angle A D B=\theta, \angle A B D=\alpha, B C=p and CD=qC D=q and CD=qC D=q.
Question: What is ABA B equal to?

Options

  1. A(p2+q2)sin⁡θpcos⁡θ+qsin⁡θ\frac{\left(p^2+q^2\right) \sin \theta}{p \cos \theta+q \sin \theta}
  2. B(p2−q2)sin⁡θpcos⁡θ+qsin⁡θ\frac{\left(p^2-q^2\right) \sin \theta}{p \cos \theta+q \sin \theta}
  3. C(p2+q2)sin⁡θqcos⁡θ+psin⁡θ\frac{\left(p^2+q^2\right) \sin \theta}{q \cos \theta+p \sin \theta}
  4. D(p2−q2)cos⁡θqcos⁡θ+psin⁡θ\frac{\left(p^2-q^2\right) \cos \theta}{q \cos \theta+p \sin \theta}

Correct answer

A. (p2+q2)sin⁡θpcos⁡θ+qsin⁡θ\frac{\left(p^2+q^2\right) \sin \theta}{p \cos \theta+q \sin \theta}

Step-by-step solution

From △ABD,\triangle A B D,ABsin⁡θ=BDsin⁡(180∘−(α+θ))⇒AB=BD⋅sin⁡θsin⁡(α+θ)=BD⋅sin⁡θsin⁡α⋅cos⁡θ+cos⁡α⋅sin⁡θ\begin{aligned} & \frac{A B}{\sin \theta}=\frac{B D}{\sin \left(180^{{}^{\circ}}-(\alpha+\theta)\right)} \\ & \Rightarrow A B=\frac{B D \cdot \sin \theta}{\sin (\alpha+\theta)}=\frac{B D \cdot \sin \theta}{\sin \alpha \cdot \cos \theta+\cos \alpha \cdot \sin \theta} \end{aligned}Again from $\triangle B C D,\begin{aligned} & B D=\sqrt{(B C)^2+(C D)^2}=\sqrt{p^2+q^2} \\ & \sin \alpha=\frac{B C}{B D}=\frac{p}{\sqrt{p^2+q^2}} \\ & \cos \alpha=\frac{C D}{B D}=\frac{q}{\sqrt{p^2+q^2}}\end{aligned}$$\begin{aligned} \therefore A B & =\frac{\sqrt{p^2+q^2} \cdot \sin \theta}{\frac{p \cdot \cos \theta}{\sqrt{p^2+q^2}}+\frac{q \cdot \sin \theta}{\sqrt{p^2+q^2}}} \\ & =\frac{\left(p^2+q^2\right) \cdot \sin \theta}{p \cdot \cos \theta+q \cdot \sin \theta}\end{aligned}
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