Krishna
0

Step 1: According to the given values construct a triangle

             CONSTRUCTION:

          


Step 2: Find the suitable trigonometric ratio to fin the unknown lengths

           NOTE:   \sin 60\degree = \frac{opposite}{hypotenuse} = \frac{QR}{PR}

                         \cos 60\degree = \frac{adjacent}{hypotenuse} = \frac{PQ}{PR}


Step 3: Find the unknown lengths

           EXAMPLE:     \cos 60\degree = \frac{PQ}{PR} = \frac{6}{PR}

                                           \frac{1}{2} = \frac{6}{PR} (since \cos 60\degree = \frac{1}{2})

                                           PR = 12 .........................(1)


                                 \sin 60\degree = \frac{QR}{PR} = \frac{QR}{12}

                                       \frac{\sqrt{3}}{2} = \frac{QR}{12}

                                     QR = \frac{12*\sqrt{3}}{2}

                                     QR = 6 \sqrt{3}


                           PR = 12 cm,   QR = 6 \sqrt{3}