Question

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what is the measure of ∠klj?
use the keypad to enter your answer in the box. include the correct symbol for the units of measure.
the measure of ∠klj is \boxed{}.

Explanation:

Step1: Identify the circle properties

The line \( QJ \) is a radius (or diameter extension) of the semicircle. \( \angle KJL \) is likely a right angle? Wait, no, actually, if \( J \) is on the semicircle and \( Q \) is the center (since \( QJ \) is a vertical line through the center of the semicircle), then \( \triangle KLJ \): Wait, maybe \( \angle KLJ \) is an inscribed angle subtended by a right angle? Wait, no, let's think again. If \( Q \) is the center, then \( QJ \) is a radius, and the arc \( KJ \) (assuming \( K \) is the right end, but maybe \( J \) is at the bottom of the semicircle, and \( Q \) is the center. Then \( \angle KLJ \): Wait, maybe the triangle is a right triangle? Wait, no, the key here is that if \( J \) is on the semicircle, and \( Q \) is the center, then \( \angle KLJ \) is 90 degrees? Wait, no, inscribed angle subtended by a diameter is 90 degrees. Wait, the semicircle has diameter, so the angle subtended by the diameter is a right angle. So if \( J \) and another point (say \( K \)) are on the semicircle, and \( L \) is another point, but maybe \( \angle KLJ \) is a right angle? Wait, maybe the diagram shows that \( QJ \) is vertical, and \( LJ \) and \( KJ \) form a triangle where \( \angle KLJ = 90^\circ \)? Wait, no, maybe the measure is \( 90^\circ \). Wait, maybe I made a mistake. Wait, the problem is about a semicircle, so the angle inscribed in a semicircle is a right angle. So if \( J \) and \( K \) are endpoints of the diameter? Wait, no, the semicircle is drawn, with \( J \) at the bottom, \( Q \) at the center (top of \( J \) on the vertical line). Then \( L \) is a point on the semicircle. Then \( \angle KLJ \): Wait, maybe \( K \) is the right end, \( J \) is the bottom, \( L \) is a point on the left. Then \( \angle KLJ \) is an angle in the triangle. Wait, maybe the answer is \( 90^\circ \). Wait, let's recall: In a circle, an angle inscribed in a semicircle is a right angle (Thales' theorem). So if \( KJ \) is the diameter, then any point \( L \) on the semicircle will form \( \angle KLJ = 90^\circ \). So assuming \( KJ \) is the diameter (since \( Q \) is the center, \( QJ \) is a radius, so \( KJ \) would be the diameter if \( K \) is on the other end), then \( \angle KLJ = 90^\circ \).

Step2: Confirm the angle measure

Using Thales' theorem, the angle subtended by a diameter in a semicircle is a right angle. So \( \angle KLJ = 90^\circ \).

Answer:

\( 90^\circ \)