Question

2. In the figure /_ PQR is a right triangle.

• See the length of the PQ
• See the length of the QR​

Answer 1

$\red{\underline{\texttt{Given:}}}$

$\sf{\overline{PR} = 30 \: cm}$

$\sf{\angle P = 30 \degree}$

$\green{\underline{\texttt{To\:find:}}}$

$\sf{\overline{PQ}, \overline{QR}}$

$\red{\underline{\texttt{Solution:}}}$

$\textsf{We know that:}$

$\begin{gathered}\sf{sin \: \theta = \frac{Opposite Side}{Hypotenuse}}\\\end{gathered}$

$\begin{gathered}\sf{ sin 30 = \frac{\overline{RQ}}{\overline{PR}}}\\\end{gathered}$

$</p><p>\begin{gathered}\sf{ \frac{1}{2} = \frac{\overline{RQ}}{12}}\\\end{gathered}$

$</p><p>\begin{gathered}\sf{ \overline{RQ} = 6 \: cm} \\\end{gathered}$

$\textsf{We also know that:}$

$\begin{gathered}\sf{cos \: \theta = \frac{Adjacent Side}{Hypotenuse}}\\\end{gathered}$

$\begin{gathered}\sf{ cos 30 = \frac{\overline{PQ}}{\overline{PR}}}\\\end{gathered}$

$\begin{gathered}\sf{ \frac{\sqrt{3}}{2} = \frac{\overline{PQ}}{12}}\\\end{gathered}$

$\sf{ \overline{PQ} = 6\sqrt{3} \: cm}$

$\textsf{Thus:}$

$\boxed{\sf{\pink{\overline{RQ} = 6\: cm, \overline{PQ} = 6\sqrt{3}\: cm}}}$

Answer 2

Answer:

$\huge{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$

Given:

$\sf{\overline{PR} = 30 \: cm}$

$\sf{\angle P = 30 \degree}$

$\green{\underline{\texttt{To\:find:}}}$

$\sf{\overline{PQ}, \overline{QR}}$

$\red{\underline{\texttt{Solution:}}}$

$\textsf{We know that:}$

$\begin{gathered}\begin{gathered}\sf{sin \: \theta = \frac{Opposite Side}{Hypotenuse}}\\\end{gathered}\end{gathered}$

$\begin{gathered} \begin{gathered}\sf{ \frac{1}{2} = \frac{\overline{RQ}}{12}}\\\end{gathered}\end{gathered} </strong></p><p><strong>[tex]\begin{gathered} \begin{gathered}\sf{ \frac{1}{2} = \frac{\overline{RQ}}{12}}\\\end{gathered}\end{gathered}$

$\begin{gathered} \begin{gathered}\sf{ \overline{RQ} = 6 \: cm} \\\end{gathered}\end{gathered}$

$\textsf{We also know that:}$

$\begin{gathered}\begin{gathered}\sf{cos \: \theta = \frac{Adjacent Side}{Hypotenuse}}\\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\sf{ cos 30 = \frac{\overline{PQ}}{\overline{PR}}}\\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\sf{ \frac{\sqrt{3}}{2} = \frac{\overline{PQ}}{12}}\\\end{gathered}\end{gathered}$

$\sf{ \overline{PQ} = 6\sqrt{3} \: cm}$

$\textsf{Thus:}$

$\boxed{\sf{\pink{\overline{RQ} = 6\: cm, \overline{PQ} = 6\sqrt{3}\: cm}}}$