what formula would i use to solve the problem for area of this triangle: triangle ABC, a=3, angleB=24, angleC=24

 triangle ABC, a=3, angleB=24, angleC=24

 

1/   You need to find the height(h) of this triangle:                                                                                                        

        

$$h = tan(B)(a/2)$$

 

 2/  Now you can calculate the area of the triangle using this formula:

 

        

$$Area(A) = h(a/2)$$

what formula would i use to solve the problem for area of this triangle: triangle ABC, a=3, angleB=24, angleC=24

$$\\(1) \qquad Area=\dfrac{a\cdot b}{2}\cdot\sin{(C)}\\\\ (2) \qquad b=a \cdot \dfrac{\sin{(C)}} {\sin{(A)}} }\\\\ (3) \qquad \sin{(A)}= \sin{( 180\ensurement{^{\circ}}-(B+C) )} = \sin{( B+C ) }$$

 

$$Area=\dfrac{a^2}{2} \cdot \dfrac{ \sin{(B)} \cdot\sin{(C)} } {\sin{(B+C)}} = \dfrac{a^2}{2} \cdot \left( \dfrac{ 1 } { \cot{(B)} + \cot{(C)} } } \right)\\\\\\ Area = \dfrac{3^2}{2} \cdot \left( \dfrac{ 1 } { \cot{(24\ensurement{^{\circ}})} + \cot{(24\ensurement{^{\circ}})} } } \right)\\\\ Area = \dfrac{3^2}{2} \cdot \left( \dfrac{ 1 } { 2\cdot \cot{(24\ensurement{^{\circ}})} } } \right)\\\\ \boxed{ Area = \dfrac{3^2}{4} \cdot \tan{(24\ensurement{^{\circ}})} } } \\\\ Area = \dfrac{3^2}{4} \cdot 0.4452286853\\\\ Area = \dfrac{9}{4} \cdot 0.4452286853\\\\ Area = 1.0017645419$$

civonamzuk has spotted that this is an isosceles triangle, which simplifies the calculation significantly!

.