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Mathematics Question on Trigonometric Functions

ABCDABCD is a trapezium such that ABAB and CDCD are parallel and BCBC ⊥\perp CDCD, if ∠\angle ADB=θ,BC=pADB = \theta, BC = p and CD=qCD = q, then ABAB is equal to

A

(p2+q2) sin θp cos θ+q sin θ \frac{ (p^2 + q^2 ) \, sin \, \theta}{ p \, cos \, \theta + q \, sin \, \theta}

B

p2+q2 cos θp cos θ+q sin θ \frac{ p^2 + q^2 \, cos \, \theta}{ p \, cos \, \theta + q \, sin \, \theta}

C

p2+q2p2 cos θ+q2 sin θ \frac{ p^2 + q^2}{ p^2 \, cos \, \theta + q^2 \, sin \, \theta}

D

(p2+q2) sin θ(p cos θ+q sin θ)2 \frac{ (p^2 + q^2) \, sin \, \theta}{ (p \, cos \, \theta + q \, sin \, \theta)^2}

Answer

(p2+q2) sin θp cos θ+q sin θ \frac{ (p^2 + q^2 ) \, sin \, \theta}{ p \, cos \, \theta + q \, sin \, \theta}

Explanation

Solution

Let AB = x
In △DAM\triangle DAM, tan (π−heta−α)=px−q (\pi - heta - \alpha) = \frac{p}{ x - q}
⇒tan(heta+α)=pq−x\Rightarrow tan ( heta + \alpha ) = \frac{p}{ q - x}
⇒q−x p cot(heta+α)\Rightarrow q - x \, p \, cot ( heta + \alpha)
⇒x=q−p cotheta+α)\Rightarrow x = q - p \, cot heta + \alpha)
= q - p (cot heta cot α−1cot α+cot heta)\bigg( \frac{ cot \, heta \, cot \, \alpha - 1}{ cot \, \alpha + cot \, heta } \bigg) \hspace25mm [∵ cot α=qp]\Bigg [ \because \, cot \, \alpha = \frac{q}{p} \Bigg ]
= q - p (qpcot heta−1qp+cot heta)=q−p(q cot heta−pq+pcot heta) \Bigg ( \frac{ \frac{q}{p} cot \, heta - 1 }{ \frac{q}{p} + cot \, heta }\Bigg ) = q - p \bigg( \frac{ q \, cot \, heta - p }{ q + p cot \, heta } \bigg)
= q - p (q cos heta−psin hetaq sin heta+p cos heta)\bigg ( \frac{ q \, cos \, heta - p sin \, heta }{ q \, sin \, heta + p \, cos \, heta }\bigg)
⇒x=q2 sin heta+pq cos heta−pq cos heta+p2 sin hetap cos heta+q sin heta\Rightarrow x = \frac{ q^2 \, sin \, heta + pq \, cos \, heta - pq \, cos \, heta + p^2 \, sin \, heta }{ p \, cos \, heta + q \, sin \, heta }
⇒AB=(p2+q2) sin hetap cos heta+q sin heta\Rightarrow AB = \frac{(p^2 + q^2) \, sin \, heta }{ p \, cos \, heta + q \, sin \, heta }
Alternate Solution
Applying sine rule in △ABD\triangle ABD,
ABsin heta=p2+q2sin π−(heta+α)\frac{AB}{sin \, heta } = \frac{ \sqrt{ p^2 + q^2}}{ sin \, \\{ \pi - ( heta + \alpha) \\}}
⇒ABsin heta=p2+q2sin (heta+α)\Rightarrow \frac{ AB}{ sin \, heta } = \frac{ \sqrt { p^2 + q^2 }}{ sin \, ( heta + \alpha)}
⇒AB=p2+q2sin hetacosα+coshetasinα[∵cos α=qp2+q2]\Rightarrow AB = \frac{ \sqrt{ p^2 + q^2 }}{ sin \, heta cos \alpha + cos heta sin \alpha } \bigg [ \because cos \, \alpha = \frac{ q}{ \sqrt{p^2 + q^2}} \bigg ]
= (p2+q2) sin ?p cos ?+q sin ?=pp2+q2\frac{(p^2 + q^2 ) \, sin \, ? }{ p \, cos \, ? + q \, sin \, ?} = \frac{p}{\sqrt{ p^2 + q^2 }}