Question

B. Find the area of A ABC, if ZA = 60° and
B = 30° and CB=6 cm.​

Answer 1

Answer:

Area of triangle is

$2 \sqrt{3 } \times 6 { \:cm}^{ 2}$

Step-by-step explanation:

This can be obtained by finding AC using trigonometry and finding the area of triangle by ½×b×h.

Answer 2

Given:

In a$\Delta ABC$, $BC = 6cm,$ $\angle B = 30^\circ$ and $\angle A = 60^\circ$

To Find:

• Area of  $\Delta ABC$

Solution:

       We know that,

           In a triangle sum of all angles are $180^ \circ$

            In $\Delta ABC$

                 $\angle A + \angle B+ \angle C = 180^\circ\\ \angle C = 180^\circ - \angle A - \angle B$

                 $\angle C = 180^\circ - 30^\circ - 60^\circ$

                $\angle C = 90^\circ$

        Thus,  $\Delta ABC$ is a right angle triangle

          Image of $\Delta ABC$ is attached below

                     $tan B = \frac{AC}{CB}$

                     $tan 30^\circ = \frac{AC}{6}$

                          $AC =\frac{6}{\sqrt{3} } \\AC = 2\sqrt{3}$

                Area of $\Delta ABC$ $= \frac{1}{2} \times CB \times AC$

                                         $= \frac{1}{2} \times 6 \times 2\sqrt{3}$

                                          $= 6\sqrt{3} cm^2$

Thus, Area of $\Delta ABC$   $= 6\sqrt{3} cm^2$