QUESTION IMAGE
Question
in △jkl, the measures of both ∠j and ∠k are equal and the measure of ∠l is 148°. what is the measure of ∠j? a 90° b 74° c 32° d 16°
Step1: Recall triangle angle sum
The sum of angles in a triangle is \(180^\circ\). So, \(\angle J+\angle K+\angle L = 180^\circ\).
Step2: Let \(\angle J = \angle K = x\)
Given \(\angle L = 148^\circ\), substitute into the formula: \(x + x+ 148^\circ= 180^\circ\).
Step3: Solve for \(x\)
Combine like terms: \(2x+ 148^\circ= 180^\circ\). Subtract \(148^\circ\) from both sides: \(2x = 180^\circ - 148^\circ= 32^\circ\). Divide by 2: \(x=\frac{32^\circ}{2}= 16^\circ\)? Wait, no, wait. Wait, \(180 - 148 = 32\), then \(32\) divided by 2 is \(16\)? Wait, no, wait, I made a mistake. Wait, \(180 - 148 = 32\)? No, \(180 - 148 = 32\)? Wait, \(148 + 32 = 180\), yes. Then \(2x = 32\), so \(x = 16\)? Wait, no, the options have 16, 32, 74, 90. Wait, maybe I messed up. Wait, no, wait, \(\angle L\) is \(148^\circ\), so the other two angles sum to \(180 - 148 = 32^\circ\). Since they are equal, each is \(32/2 = 16^\circ\)? But option D is \(16^\circ\). Wait, but let me check again. Wait, maybe I miscalculated. \(180 - 148 = 32\), yes. Then two equal angles, so each is \(16\). So \(\angle J = 16^\circ\)? But wait, the options: D is \(16^\circ\). Wait, but let me check the problem again. "the measures of both \(\angle J\) and \(\angle K\) are equal and the measure of \(\angle L\) is \(148^\circ\)". So triangle angle sum: \(J + K + L = 180\). \(J = K\), so \(2J + 148 = 180\). \(2J = 180 - 148 = 32\). \(J = 16\). So the answer is D? Wait, but the options: D is \(16^\circ\). Wait, but I thought maybe I made a mistake, but the calculation seems right.
Wait, no, wait, maybe I misread the angle. Wait, is \(\angle L\) the vertex angle? Let me confirm. In triangle \(JKL\), angles at \(J\) and \(K\) are equal, so it's an isosceles triangle with \(J = K\), and \(L\) is the vertex angle. So sum of angles: \(J + K + L = 180\). \(J = K\), so \(2J = 180 - L\). \(L = 148\), so \(2J = 32\), \(J = 16\). So the answer is D.
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D. \(16^\circ\)