A, B and C form the vertices of a triangle. BC = 5.7cm and angle CBA = 137 degrees. Given that the area of the triangle is  10.4 cm^2, what is the perimeter of the triangle?

Jul 31, 2018

$21.33 \text{cm (2dp)}$.

Explanation:

Recall that, the Area of $\Delta A B C = \frac{1}{2} \cdot B C \cdot B A \cdot \sin \angle C B A$.

$\therefore 10.4 = \left(\frac{1}{2}\right) \left(5.7\right) \left(B A\right) \sin {137}^{\circ} ,$

$i . e . , 10.4 = 2.85 \cdot B A \cdot \left(0.682\right) = \left(1.9437\right) B A$.

$\therefore B A = \frac{10.4}{1.9437} = 5.35 \ldots \ldots \ldots \ldots \ldots \left(1\right)$.

Further, by the cosine-rule,

$A {C}^{2} = C {B}^{2} + B {A}^{2} - 2 \cdot C B \cdot B A \cdot \cos \angle C B A$,

$= {5.7}^{2} + {5.35}^{2} - 2 \left(5.7\right) \left(5.35\right) \left(- 0.731\right)$,

$= 32.49 + 28.62 + 44.58 ,$

$A {C}^{2} = 105.69$.

$\therefore A C = \sqrt{105.69} = 10.28 \ldots \ldots \ldots \ldots \ldots \left(2\right)$.

Thus, we have, $B A = 5.35 , A C = 10.28 , C B = 5.7$.

Hence, the reqd. perimeter$= 5.35 + 10.28 + 5.7 = 21.33 c m$.