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

Q: Wait, why do we need to draw the diagonal $BD$? Can't we just use coordinate geometry?

You absolutely could use coordinate geometry, but it would be like using a sledgehammer to crack a nut! Drawing the diagonal $BD$ is a classic geometric strategy. It creates a bridge between the knowns ($p, q$) and the unknown ($AB$). By splitting the trapezium into two triangles, we can use the Sine Rule, which is much more elegant and less prone to algebraic errors than setting up coordinates for every vertex.

Q: Why is $\angle ABD = \angle BDC$? I always get confused with parallel line properties.

That is a common hurdle! Remember the 'Z' shape. When you have two parallel lines ($AB$ and $CD$) and a transversal ($BD$), the alternate interior angles are always equal. Imagine tracing the letter 'Z' starting from $A$, going to $B$, then down to $D$, and finally to $C$. The angles tucked inside the corners of that 'Z' are your alternate interior angles. It is a powerful tool to transfer information across the figure.

Q: Why did we substitute $BD^2 = p^2 + q^2$ at the end? Is it necessary?

It is absolutely necessary! The problem asks for $AB$ in terms of $p, q,$ and $\theta$. Our intermediate steps introduced $BD$, which is not part of the final answer. By using Pythagoras' theorem in $\triangle BCD$, we express $BD$ in terms of $p$ and $q$. Substituting $BD^2 = p^2 + q^2$ is the final step to clean up our expression and ensure our answer is in the required format.

Question

$ABCD$ is a trapezium such that $AB$ and $CD$ are parallel and $BC \perp CD$. If $\angle ADB = 	heta, BC = p$ and $CD = q$, then $AB$ is equal to

Options:

(A) $\frac{(p^2+q^2) \sin 	heta}{p \cos 	heta + q \sin 	heta}$

(B) $\frac{p^2+q^2 \cos 	heta}{p \cos 	heta + q \sin 	heta}$

(C) $\frac{p^2+q^2}{p^2 \cos 	heta + q^2 \sin 	heta}$

(D) $\frac{(p^2+q^2) \sin 	heta}{(p \cos 	heta + q \sin 	heta)^2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a trapezium $ABCD$, where $AB \parallel CD$. We are given that $BC \perp CD$, which anchors our shape with a perfect ninety-degree corner at vertex $C$.

We are given the dimensions $BC = p$ and $CD = q$. This serves as our sturdy foundation for the geometric derivation.

### The Diagonal: Our Geometric Bridge

To unlock the secrets of this shape, we draw the diagonal $BD$. This line splits our trapezium into two distinct worlds: the right-angled triangle $	riangle BCD$ and the triangle $	riangle ABD$.

In $	riangle BCD$, we have a right angle at $C$. By Pythagoras' theorem, the hypotenuse $BD$ is calculated as:

$$BD = \sqrt{p^2 + q^2}$$

This diagonal acts as our bridge, carrying the information from the known lengths $p$ and $q$ into the territory of $	riangle ABD$.

### The Parallel Symmetry

Since $AB \parallel CD$ and $BD$ is a transversal, the alternate interior angles must be equal. This implies $\angle ABD = \angle BDC$. Let us define this angle as $\alpha$.

In our right-angled triangle $	riangle BCD$, we define the trigonometric ratios as follows:

$$\sin \alpha = \frac{p}{BD}, \quad \cos \alpha = \frac{q}{BD}$$

We have now successfully captured the essence of the right triangle in terms of $p$, $q$, and $BD$.

### The Sine Rule: The Master Key

Now, we shift our focus to $	riangle ABD$. We are given $\angle ADB = 	heta$, and we previously discovered $\angle ABD = \alpha$.

Since the sum of angles in a triangle is $180^\circ$, the third angle is $\angle DAB = 180^\circ - (	heta + \alpha)$. Applying the Sine Rule:

$$\frac{AB}{\sin 	heta} = \frac{BD}{\sin(180^\circ - (	heta + \alpha))}$$

Using the trigonometric identity $\sin(180^\circ - x) = \sin x$, the denominator simplifies to $\sin(	heta + \alpha)$.

### The Final Synthesis

Expanding $\sin(	heta + \alpha)$ using the addition formula gives $\sin 	heta \cos \alpha + \cos 	heta \sin \alpha$. Substituting our earlier ratios for $\sin \alpha$ and $\cos \alpha$, we obtain:

$$AB = \frac{BD \sin 	heta}{\sin 	heta (\frac{q}{BD}) + \cos 	heta (\frac{p}{BD})}$$

Multiplying the numerator and denominator by $BD$ clears the fractions:

$$AB = \frac{BD^2 \sin 	heta}{q \sin 	heta + p \cos 	heta}$$

Finally, substituting $BD^2 = p^2 + q^2$, we arrive at the elegant result:

**$$AB = \frac{(p^2 + q^2) \sin 	heta}{p \cos 	heta + q \sin 	heta}$$**

We have traversed the geometry, bridged the triangles, and arrived at the truth. Mathematics is truly a beautiful language.

Animated Solution for Mathematics - Trigonometry: $A B C D$ is a trapezium such that $A B$ and $C D$ are parallel and $B C ⊥ C D$ . If $∠ A D B = θ, B C = p$ and $C D = q$ , then $A B$ is equal to

Select Answer:

Visualized Solution

Visualizing Trapezium $A B C D$

Drawing the Diagonal $B D$

Trigonometric Ratios in $△ B C D$

Using Parallel Lines: $A B ∥ C D$

Angles of Triangle $A B D$

Applying the Sine Rule in $△ A B D$

Simplifying the Angle Expression

Substituting $sin α$ and $cos α$

The Final Elegant Expression

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Diagonal

The Parallel Symmetry

The Sine Rule

The Final Synthesis

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 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

Select Answer:

Visualized Solution

Visualizing Trapezium .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABCD

Drawing the Diagonal .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BD

Trigonometric Ratios in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △BCD

Using Parallel Lines: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB∥CD

Angles of Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABD

Applying the Sine Rule in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ABD