We are given a diagram where $\angle ABC$ and $\angle BCD$ are right angles. We are also given that $\angle BAD = t$ and $\angle EDF = 70^\circ$. The goal is to find the value of $t$.

1. Problem Description

We are given a diagram where $\angle ABC$ and $\angle BCD$ are right angles. We are also given that $\angle BAD = t$ and $\angle EDF = 70^\circ$. The goal is to find the value of $t$.

2. Solution Steps

Let's denote $\angle ADE$ as $x$. Since $\angle EDF = 70^\circ$, and $\angle ADE$ and $\angle EDF$ form a straight line, they are supplementary angles. Therefore, their sum is $180^\circ$.

$$x + 70^\circ = 180^\circ$$

$$x = 180^\circ - 70^\circ = 110^\circ$$

So, $\angle ADE = 110^\circ$.

Since $\angle ABC = 90^\circ$ and $\angle BCD = 90^\circ$, $AB$ is parallel to $CD$.

Therefore, $\angle ADC$ and $\angle BAD$ are supplementary angles.

$$t + \angle ADC = 180^\circ$$

In quadrilateral $ABCD$, the sum of the interior angles is $360^\circ$.

$$\angle ABC + \angle BCD + \angle CDA + \angle DAB = 360^\circ$$

$$90^\circ + 90^\circ + \angle CDA + t = 360^\circ$$

$$180^\circ + \angle CDA + t = 360^\circ$$

$$\angle CDA + t = 180^\circ$$

Thus $\angle CDA = 180^\circ - t$.

Since $\angle ADE$ is an exterior angle of $\angle CDA$, we have:

$$\angle ADE = 110^\circ$$

$$\angle CDA + \angle ADE = 360^\circ$$

This is not helpful.

Since AB is parallel to CD, $\angle BAD + \angle ADC = 180^{\circ}$. Then, $\angle ADC = 180^{\circ} - t$. Also, $\angle ADE = 110^{\circ}$.

$\angle ADC + \angle ADE + \angle EDF = 360$.

Consider the line CD extended to E.

Since AB is parallel to CD, we know that the two lines make the same angle with the transversal AD.

So the exterior angle at D which is $\angle ADE$ must be equal to $180 - t$ plus some correction.

So $\angle ADC = 180 - t$.

We know that the $\angle ADE$ is supplementary to $70^{\circ}$ so $\angle ADE = 110^{\circ}$. Also $\angle ADE$ and $\angle ADC$ make up $\angle CDE = 180^{\circ}$.

So $\angle ADE + \angle ADC = 360$ isn't correct.

Then $\angle CDA$ and $\angle ADE$ sum to a straight line, $\angle CDE$, meaning that it must be $180^{\circ}$. Since $\angle ADE = 110^{\circ}$, we have $\angle CDA = 180^{\circ} - 110^{\circ} = 70^{\circ}$.

Then $t + 70^{\circ} = 180^{\circ}$ since $\angle BAD$ and $\angle ADC$ are supplementary. Thus $t = 180^{\circ} - 70^{\circ} = 110^{\circ}$.

3. Final Answer

D. 110°