Question

2. In the figure ▲ PQR is a right triangle.

a) See the length of the PQ
b) See the length of the QR

Answer 1

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

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

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

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

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

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

$\textsf{We know that:}$

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

→ $\sf{ sin 30 = \frac{\overline{RQ}}{\overline{PR}}}$

→ $\sf{ \frac{1}{2} = \frac{\overline{RQ}}{12}}$

→ $\sf{ \overline{RQ} = 6 \: cm}$

$\textsf{We also know that:}$

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

→ $\sf{ cos 30 = \frac{\overline{PQ}}{\overline{PR}}}$

→ $\sf{ \frac{\sqrt{3}}{2} = \frac{\overline{PQ}}{12}}$

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

$\textsf{Thus:}$

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

Answer 2

$\huge\bf{Question}$

In the figure ▲ PQR is a right triangle.

a) See the length of the PQ

b) See the length of the QR

$\huge\bf{Answer}$

Given:-

• PR = 12cm
• ∠P = 30°

To Find:-

• Side PQ and QR

We know that,

$\large \bf {sin\: \theta = \frac{Opposite\: Side}{Hypotenuse} }$

$\large \bf {sin30° = \frac{RQ}{RP} }$

$\large \bf \frac{1}{2} = \frac{RQ}{12}$

$\large \bf \not{12} = \not {2}RQ$

$\large \bf \green {==》RQ = 6 cm}$

We also know that,

$\large \bf {sin\: \theta = \frac{Adjacent\: Side}{Hypotenuse} }$

$\large \bf {cos30° = \frac{PQ}{RP} }$

$\large \bf \frac {\sqrt{3}}{2} = \frac{PQ}{12}$

$\large \bf \not{12} {\sqrt{3}} = \not{2}PQ$

$\large \bf \green {==》PQ = 6{\sqrt{3}}cm}$