In trapezoid $ABCD$, $AD\parallel BC$, $\angle B=60^\circ$, $\angle C=45^\circ$, $AD=4$, $BC=10$. Find $AB$ and $CD$.

Step 1: Draw Perpendiculars and Identify Triangles

To solve this problem, we draw perpendiculars from vertices $A$ and $D$ to the base $BC$. Let these perpendiculars meet $BC$ at points $E$ and $F$ respectively. This creates two right-angled triangles, $\triangle ABE$ and $\triangle DFC$, and a rectangle $AEFD$.

Step 2: Determine Lengths of Segments on BC

Since $AEFD$ is a rectangle, $EF = AD = 4$. We know that $BC = 10$, and $BC$ can be expressed as the sum of $BE$, $EF$, and $FC$. Substituting the known values, we find that the sum of $BE$ and $FC$ is $6$.

Step 3: Calculate AE and BE using Trigonometry in Triangle ABE

In the right-angled triangle $\triangle ABE$, we are given $\angle B = 60^\circ$. We can use trigonometric ratios to relate $AE$, $BE$, and $AB$. Specifically, $\tan 60^\circ = \frac{AE}{BE}$ and $\sin 60^\circ = \frac{AE}{AB}$.

Step 4: Calculate DF and FC using Trigonometry in Triangle DFC

Since $AEFD$ is a rectangle, $AE = DF$. In the right-angled triangle $\triangle DFC$, we have $\angle C = 45^\circ$. Using the tangent ratio, $\tan 45^\circ = \frac{DF}{FC}$. Since $\tan 45^\circ = 1$, we get $DF = FC$.

Step 5: Solve for BE, FC, AE, AB, and CD

From the previous steps, we have $AE = BE\sqrt{3}$ and $AE = FC$. This implies $FC = BE\sqrt{3}$. We also know $BE + FC = 6$. Substituting $FC$ in the sum equation, we can solve for $BE$. Once $BE$ is found, we can find $FC$, $AE$, $AB$, and $CD$ using the trigonometric relations.

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