Question

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

1.) See the length of the PQ
2.) See the length of the QR​

Answer 1

$\huge{\mathcal {\purple {A}\green {n}\pink {s}\blue {w}\purple{e}\green {r}}}$

1. ☆cos30°=pq/12

pQ=6cm✔✔✔

2.RQ^2=+PQ^2=RP^2

QR=11CM

$\huge\colorbox{orange}{çhű࿐ ❤}$

Answer 2

$\huge\bf{Question}$

In the figure PQR is a right triangle.

1.) See the length of the PQ

2.) See the length of the QR

$\huge\bf{Answer}$

$\underline{\boxed{\pink{\sf Given:- }}}$

• PR = 12cm
• ∠P = 30°

$\underline{\boxed{\purple{\sf To \: \: find:- \: }}}$

• Side PQ and QR

$\underline{\boxed{\pink{\sf We \: know \: \: that:-}}}$

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

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

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

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

$\large \bf \green {==》RQ = 6 cm}$

$\underline{\boxed{\red{\sf We \: also \: know \: that :-}}}$

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

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

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

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

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