In a right-angled triangle ABC, with the right angle at BAC, when lines AB and AC are extended, the exterior angle A' is 37 degrees. The angle denoted as C, or t, is found to be -53 degrees, indicating a measurement opposite to the chosen reference direction.
In a right-angled triangle ABC, let \( \angle BAC \) be the right angle. If line AB and AC are extended, the exterior angle at A is denoted as \( \angle A' \).
Given that \( \angle A' = 37^\circ \), and we denote \( \angle CAB \) as \( t \), we can use the property that the exterior angle of a triangle is equal to the sum of its two opposite interior angles.
\[ \angle A' = \angle BAC + \angle CAB \]
Substitute the given values:
\[ 37^\circ = 90^\circ + t \]
Now, solve for \( t \):
\[ t = 37^\circ - 90^\circ \]
\[ t = -53^\circ \]
Therefore, the angle \( \angle C \), denoted as \( t \), is \( -53^\circ \). Note that angles can be negative when measured in a direction opposite to the chosen reference direction. In this case, the negative sign indicates that \( \angle CAB \) is measured in the direction opposite to the exterior angle \( \angle A' \).