Question

(b) find m∠k.

m∠k=□°

Explanation:

Step1: Identify Triangle Type

In triangle \( KJL \), sides \( KJ \) and \( KL \) are both 10 units, so it's isosceles with \( \angle K \cong \angle L \)? Wait, no—wait, in an isosceles triangle, the angles opposite equal sides are equal. Wait, \( KJ = KL = 10 \), so the sides opposite angles \( L \) and \( J \)? Wait, no: side \( KJ \) is opposite angle \( L \), and side \( KL \) is opposite angle \( J \). Wait, no, let's label: vertices \( K \), \( J \), \( L \). Side \( KJ \) is between \( K \) and \( J \), length 10. Side \( KL \) is between \( K \) and \( L \), length 10. So the two equal sides are \( KJ \) and \( KL \), so the base is \( JL \). Therefore, the base angles are \( \angle J \) and \( \angle L \)? Wait, no, wait: in triangle, equal sides are \( KJ \) (from \( K \) to \( J \)) and \( KL \) (from \( K \) to \( L \)), so the angles opposite those sides: side \( KJ \) is opposite angle \( L \), side \( KL \) is opposite angle \( J \). Wait, no, side opposite \( \angle K \) is \( JL \), side opposite \( \angle J \) is \( KL \), side opposite \( \angle L \) is \( KJ \). Since \( KJ = KL = 10 \), then \( \angle L = \angle J \)? Wait, no, \( KJ = KL \), so side \( KJ \) (opposite \( \angle L \)) and side \( KL \) (opposite \( \angle J \)) are equal, so \( \angle L = \angle J \). Wait, but \( \angle J \) is given as \( 70^\circ \). Wait, that can't be. Wait, maybe I mixed up. Wait, the triangle has \( KJ = 10 \), \( KL = 10 \), so the two equal sides are \( KJ \) and \( KL \), so the angles at \( J \) and \( L \) are equal? Wait, no, the angles at the base (the third side) are equal. Wait, maybe I made a mistake. Wait, let's re-express: in triangle \( KJL \), \( KJ = KL = 10 \), so the triangle is isosceles with \( K \) as the vertex? No, \( K \) is the common vertex of the two equal sides, so the base is \( JL \), and the base angles are \( \angle J \) and \( \angle L \). Wait, but \( \angle J \) is given as \( 70^\circ \). Wait, no, the problem says \( \angle J \) is \( 70^\circ \). Wait, maybe I got the equal sides wrong. Wait, looking at the diagram: \( KJ \) is from \( K \) to \( J \), length 10. \( KL \) is from \( K \) to \( L \), length 10. So sides \( KJ \) and \( KL \) are equal, so angles opposite them: angle opposite \( KJ \) is \( \angle L \), angle opposite \( KL \) is \( \angle J \). Therefore, \( \angle L = \angle J \). But \( \angle J \) is \( 70^\circ \), so \( \angle L = 70^\circ \). Then, the sum of angles in a triangle is \( 180^\circ \), so \( \angle K + \angle J + \angle L = 180^\circ \). So \( \angle K + 70^\circ + 70^\circ = 180^\circ \). Therefore, \( \angle K = 180 - 140 = 40^\circ \). Wait, that makes sense. Wait, maybe I confused the equal sides. Let's correct: if two sides are equal ( \( KJ = KL \) ), then the angles opposite those sides are equal. Side \( KJ \) is opposite \( \angle L \), side \( KL \) is opposite \( \angle J \). So \( \angle L = \angle J = 70^\circ \). Then, \( \angle K = 180 - 70 - 70 = 40^\circ \).

Step2: Calculate \( \angle K \)

Sum of angles in a triangle: \( m\angle K + m\angle J + m\angle L = 180^\circ \).
Given \( m\angle J = 70^\circ \), and \( \angle L = \angle J \) (since \( KJ = KL \), isosceles triangle), so \( m\angle L = 70^\circ \).
Thus, \( m\angle K = 180 - 70 - 70 = 40^\circ \).

Answer:

\( 40 \)