ABCD is a trapezium such that AB and CD are parallel and BC CD . If ADB = θ , BC = p and CD = q , then AB is equal to: - Mathematics MCQ JEE Main 2026 | ExamTest.live

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ABCD is a trapezium such that AB and CD are parallel and BC ⊥ CD . If ∠ ADB = θ, BC = p and CD = q, then AB is equal to:

Solution & Step-by-step Explanation

Let ∠ BDC = α . In △ BDC, tan α = \frac{p}{q}, so sin α = \frac{p}{p ^{2} + q ^{2}} and cos α = \frac{q}{p ^{2} + q ^{2}} .In △ ABD, using the Sine Rule: \frac{AB}{sin θ} = \frac{BD}{sin ( ∠ DAB )} Since AB ∥ CD, ∠ ABD = α . Thus ∠ DAB = π - (θ + α) . AB = \frac{BD sin θ}{sin ( π - ( θ + α ))} = \frac{p ^{2} + q ^{2} sin θ}{sin θ cos α + cos θ sin α} Substitute sin α and cos α : AB = \frac{p ^{2} + q ^{2} sin θ}{sin θ \cdot \frac{q}{p ^{2} + q ^{2}} + cos θ \cdot \frac{p}{p ^{2} + q ^{2}}} AB = \frac{( p ^{2} + q ^{2} ) sin θ}{q sin θ + p cos θ}

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ABCD is a trapezium such that AB and CD are parallel and BC ⊥ CD . If ∠ ADB = θ, BC = p and CD = q, then AB is equal to:

\frac{p ^{2} + q ^{2} c o s θ}{p c o s θ + q s i n θ}

\frac{p ^{2} + q ^{2}}{p c o s θ + q s i n θ}

\frac{( p ^{2} + q ^{2} ) s i n θ}{p c o s θ + q s i n θ}

\frac{( p ^{2} + q ^{2} ) s i n θ}{q c o s θ + p s i n θ}

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