Question

1. what is the measure of y (angle axb)?

options: 90°, 60°, 30°, 45°

Explanation:

Step1: Find AY in right △BYZ

In right triangle $BYZ$, $\tan(60^\circ)=\frac{AY}{YZ}$.
$\tan(60^\circ)=\sqrt{3}$, $YZ=15$ cm.
$AY = 15\times\tan(60^\circ)=15\sqrt{3}$ cm

Step2: Find AB in right △ABY

In right triangle $ABY$, $\angle ABY=45^\circ$, so it is an isosceles right triangle: $AB=AY=15\sqrt{3}$ cm.

Step3: Find XY in right △XYZ

In right triangle $XYZ$, $\tan(60^\circ)=\frac{XY}{YZ}$, so $XY=15\sqrt{3}$ cm.

Step4: Identify △XAB properties

$XA=XY-AY=15\sqrt{3}-15\sqrt{3}=0$? Correction: $XA=XY - AY$ is wrong. Correct: $XY$ is the vertical side, $AY=AB=15$ cm (correction: in △ABY, $\angle ABY=45^\circ$, right angle at A, so $AY=AB$. In △BYZ, $\cos(60^\circ)=\frac{YZ}{BZ}$, $BZ=30$, $BY=15\sqrt{3}$. So $AY=AB=15$ cm, $XY=15\sqrt{3}$ cm, so $XA=XY - AY=15(\sqrt{3}-1)$ cm.

Step5: Calculate $\tan(y)$ in △XAB

$\tan(y)=\frac{AB}{XA}=\frac{15}{15(\sqrt{3}-1)}=\frac{1}{\sqrt{3}-1}=\frac{\sqrt{3}+1}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{\sqrt{3}+1}{2}$
This equals $\tan(75^\circ)$? No, correction: $\tan(15^\circ)=2-\sqrt{3}$, $\tan(30^\circ)=\frac{1}{\sqrt{3}}$. Wait, correct approach: $\angle XYB=90^\circ$, $\angle ZBY=30^\circ$, $\angle XBA=180-45=135^\circ$, $\angle BXY=y$, $\angle XBY=90-30=60^\circ$. In △XBY, $\angle XYB=90^\circ$, $\angle XBY=60^\circ$, so $\angle BXY=30^\circ$.

Answer:

30°