Question

triangle wxy is isosceles. ∠ywx and ∠yxw are the base angles. yz bisects ∠wyx. m∠xyz=(15x)°. m∠yxz=(2x + 5)°. what is the measure of ∠wyx? 5° 15° 75° 150°

Explanation:

Step1: Identify triangle properties

In isosceles triangle \( WXY \), base angles \( \angle YWX \) and \( \angle YXW \) are equal. \( YZ \) bisects \( \angle WYX \), so \( \angle XYZ=\angle WYZ \). Also, in triangle \( XYZ \), we can use angle - sum property or the fact that \( \angle YXZ \) is a base angle and \( \angle XYZ \) is related to the vertex angle. Since \( \angle YXW \) is a base angle and \( \angle YXZ \) is part of it? Wait, no. Wait, \( \angle YXW \) is a base angle, and \( YZ \) is a bisector of the vertex angle \( \angle WYX \). Also, in triangle \( XYZ \), we know that \( \angle YXZ \) is a base angle - related angle? Wait, actually, since \( \angle YWX=\angle YXW \), and \( \angle YXZ \) is \( \angle YXW \), and \( \angle XYZ \) is half of \( \angle WYX \). Also, in triangle \( XYZ \), the sum of angles is \( 180^{\circ} \), but wait, actually, since \( \angle YXZ \) is a base angle and \( \angle XYZ \) is an angle formed by the bisector, and since the triangle is isosceles with base angles equal, and the bisector of the vertex angle is also the altitude and median (in an isosceles triangle, the angle bisector of the vertex angle is perpendicular to the base? Wait, no, only in an equilateral triangle or if it's also the median and altitude. Wait, no, in an isosceles triangle, the angle bisector of the vertex angle is also the median and altitude to the base. So \( YZ \) is perpendicular to \( WX \), so \( \angle YZX = 90^{\circ} \). Therefore, in triangle \( XYZ \), \( \angle XYZ+\angle YXZ + \angle YZX=180^{\circ} \), and \( \angle YZX = 90^{\circ} \), so \( \angle XYZ+\angle YXZ=90^{\circ} \).

Step2: Set up the equation

We know that \( m\angle XYZ=(15x)^{\circ} \) and \( m\angle YXZ=(2x + 5)^{\circ} \), and \( \angle XYZ+\angle YXZ = 90^{\circ} \) (because \( YZ\perp WX \) as it is the angle - bisector, median and altitude in isosceles triangle \( WXY \)). So we set up the equation:
\( 15x+(2x + 5)=90 \)

Step3: Solve for \( x \)

Combine like terms: \( 15x+2x+5 = 90\)
\( 17x+5=90 \)
Subtract 5 from both sides: \( 17x=90 - 5=85 \)
Divide both sides by 17: \( x=\frac{85}{17}=5 \)

Step4: Find \( m\angle XYZ \)

Now that \( x = 5 \), \( m\angle XYZ=15x=15\times5 = 75^{\circ} \)? Wait, no, wait. Wait, \( \angle XYZ \) is half of \( \angle WYX \), because \( YZ \) bisects \( \angle WYX \). Wait, no, earlier mistake. Wait, if \( YZ \) is the angle bisector of \( \angle WYX \), then \( \angle WYX = 2\angle XYZ \). And from the right triangle \( XYZ \) (since \( YZ\perp WX \)), \( \angle XYZ+\angle YXZ=90^{\circ} \). We found \( x = 5 \), so \( \angle XYZ=15x = 75^{\circ} \)? No, that can't be, because then \( \angle YXZ=2x + 5=15^{\circ} \), and \( 75 + 15=90 \), which is correct for the right triangle. Then \( \angle WYX=2\angle XYZ \)? Wait, no, wait, \( \angle XYZ \) is half of \( \angle WYX \), so \( \angle WYX = 2\times\angle XYZ \). Wait, no, if \( YZ \) bisects \( \angle WYX \), then \( \angle WYX=2\angle XYZ \). But we found \( \angle XYZ = 75^{\circ} \), then \( \angle WYX=150^{\circ} \)? But that can't be, because the sum of angles in a triangle is \( 180^{\circ} \), and if the vertex angle is \( 150^{\circ} \), the base angles would be \( \frac{180 - 150}{2}=15^{\circ} \), which matches \( \angle YXZ = 15^{\circ} \) (since \( 2x+5=2\times5 + 5 = 15^{\circ} \)). Wait, let's re - check.

Wait, the base angles are \( \angle YWX=\angle YXW = 15^{\circ} \) (since \( 2x + 5=15^{\circ} \) when \( x = 5 \)). The vertex angle \( \angle WYX=180-2\times15=150^{\circ} \), and \( YZ \) bisects \( \angle WYX \), so \( \angle XYZ=\frac{150}{2}=75^{\circ} \), and \( 15x=15\times5 = 75^{\circ} \), which matches. So the measure of \( \angle WYX \) is \( 150^{\circ} \)? Wait, no, wait the options are \( 5^{\circ},15^{\circ},75^{\circ},150^{\circ} \). Wait, but let's check again.

Wait, the mistake was in assuming \( YZ\perp WX \). Wait, in an isosceles triangle, the angle bisector of the vertex angle is also the median and altitude, so it is perpendicular to the base. So \( YZ\perp WX \), so triangle \( XYZ \) is right - angled at \( Z \). So \( \angle XYZ+\angle YXZ = 90^{\circ} \). So \( 15x+(2x + 5)=90 \), \( 17x=85 \), \( x = 5 \). Then \( \angle XYZ=15\times5 = 75^{\circ} \), and \( \angle WYX = 2\times75^{\circ}=150^{\circ} \)? But let's check the angle sum: \( 150 + 15+15=180 \), which is correct.

Answer:

\( 150^{\circ} \) (Option: 150°)