Question

In $△XYZ$, $\angle Y={90}^o$, $\angle Z=a^o$ and $\angle X=(a+30{)}^o$. If XZ = 24,

find XY and YZ.

• A. $XY=$ $12$ cm and $YZ=$ $12$ $\sqrt{3}$ cm
• B. $XY=$ 12$\sqrt{3}$ cm and $YZ=$ $12$ cm
• C. $XY=$ $2\sqrt{3}$ cm and $YZ=$ $12$ cm
• D. $XY=12\sqrt{3}$ cm and $YZ=$ $2$ cm

Answer 1

The correct option is A $XY=$ $12$ cm and $YZ=$ $12$ $\sqrt{3}$ cm

In $△XYZ,\angle Y={90}^o,\angle Z=a^o$ and $\angle X=(a+30{)}^o$

$XZ=24$ ...... (Given)

In triangle $XYZ,$

$\angle X+\angle Y+\angle Z={180}^o$

$⟹(a+30{)}^o+{90}^o+a^o={180}^o$

$⟹2a={60}^o$

$⟹a={30}^o$

$\therefore \angle Z={30}^o$ and $\angle Y={60}^o$

Now, $\sin {30}^o=\frac{XY}{XZ}$

$⟹\frac{1}{2}=\frac{XY}{24}$

$⟹XY=12cm$

Also, $\cos {30}^o=\frac{YZ}{XZ}$

$⟹\frac{\sqrt{3}}{2}=\frac{YZ}{24}$

$⟹YZ=12\sqrt{3}cm$