Question

given abcd is a rectangle, find ( mangle ade ).
( mangle ade=) 。

Explanation:

Step1: Identify triangle properties in rectangle

In rectangle \(ABCD\), \(AE = BE\) (diagonals of a rectangle are equal and bisect each other), so \(\triangle ABE\) is isosceles? Wait, no, actually, \(\angle BAE\) and \(\angle ABE\)? Wait, no, looking at the angles: \(\angle ABE=(4x + 15)^\circ\) and \(\angle EBC=(13x + 7)^\circ\). Since \(ABCD\) is a rectangle, \(\angle ABC = 90^\circ\), so \(\angle ABE+\angle EBC=90^\circ\).

So, \((4x + 15)+(13x + 7)=90\).

Step2: Solve for \(x\)

Combine like terms: \(4x+13x + 15 + 7=90\)
\(17x+22 = 90\)
Subtract 22 from both sides: \(17x=90 - 22=68\)
Divide by 17: \(x=\frac{68}{17}=4\).

Step3: Find \(\angle ADE\)

First, find \(\angle ADB\) or \(\angle ADE\). In rectangle, \(AD\parallel BC\), and \(AB\parallel CD\). Also, \(AE = DE\) (diagonals bisect each other), so \(\triangle ADE\) is isosceles? Wait, alternatively, \(\angle ADE=\angle DAE\), but maybe easier to find \(\angle ABE\) first. \(\angle ABE=4x + 15=4(4)+15=16 + 15=31^\circ\). Then, in rectangle, \(AD\parallel BC\), so \(\angle ADE=\angle DBC\)? Wait, no, \(\angle DBC=\angle EBC=(13x + 7)^\circ\). Wait, no, \(\angle ADE\): since \(ABCD\) is rectangle, \(AD\perp AB\)? No, \(AD\) and \(AB\) are perpendicular. Wait, diagonals in rectangle are equal, so \(AE = BE = CE = DE\). So \(\angle ADE=\angle DAE\), and \(\angle ABE=\angle BAE\) (since \(AE = BE\)). Wait, \(\angle BAE=\angle ABE = 31^\circ\), so in \(\triangle ABD\), \(\angle BAD = 90^\circ\), so \(\angle ADB=90^\circ-\angle ABD\). Wait, \(\angle ABD=\angle ABE = 31^\circ\), so \(\angle ADE=\angle ADB=90^\circ - 31^\circ=59^\circ\)? Wait, no, wait: \(\angle EBC=13x + 7=13(4)+7=52 + 7=59^\circ\). Wait, in rectangle, \(AD\parallel BC\), so \(\angle ADE=\angle DBC=\angle EBC=59^\circ\)? Wait, no, let's re-examine.

Wait, \(\angle ABC = 90^\circ\), \(\angle ABE=31^\circ\), \(\angle EBC=59^\circ\). Since \(AD\parallel BC\), \(\angle ADB=\angle DBC\) (alternate interior angles). \(\angle DBC=\angle EBC=59^\circ\), so \(\angle ADE=\angle DBC=59^\circ\)? Wait, no, \(\angle ADE\) is \(\angle ADB\), which is equal to \(\angle DBC\) because \(AD\parallel BC\) and \(BD\) is transversal. So \(\angle ADE = 59^\circ\)? Wait, let's check again.

Wait, \(x = 4\), so \(\angle EBC=13(4)+7=59^\circ\). In rectangle, \(AD\parallel BC\), so \(\angle ADE=\angle DBC=\angle EBC=59^\circ\). Alternatively, in \(\triangle ABD\), \(\angle BAD = 90^\circ\), \(\angle ABD=31^\circ\), so \(\angle ADB=90 - 31=59^\circ\). Yes, that's correct. So \(\angle ADE = 59^\circ\).

Answer:

\(59\)