QUESTION IMAGE
Question
- ( mangle ewl ) ( (7x + 5)^{circ} ) ( (3x + 21)^{circ} )
Step1: Use the Angle Bisector Theorem (HL Congruence)
Since \( LE = LJ\) (marked as equal), \( \angle WEL=\angle WJL = 90^{\circ}\), and \( WL = WL\) (common side), by the Hypotenuse - Leg (HL) congruence criterion for right - triangles \(\triangle WEL\cong\triangle WJL\). Then \( \angle EWL=\angle JWL\). So \(7x + 5=3x+21\).
Step2: Solve the equation for \(x\)
Subtract \(3x\) from both sides: \(7x-3x + 5=3x-3x + 21\), which gives \(4x+5 = 21\).
Subtract \(5\) from both sides: \(4x+5 - 5=21 - 5\), so \(4x=16\).
Divide both sides by \(4\): \(x=\frac{16}{4}=4\).
Step3: Calculate \(m\angle EWL\)
Substitute \(x = 4\) into \(7x + 5\). Then \(m\angle EWL=7\times4+5=28 + 5=33^{\circ}\).
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\(33^{\circ}\)