Question

Area do triângulo, por favor.

Answer 1

$\Large\boxed{\begin{array}{l}\underline{\rm\acute Area~do~tri\hat angulo~em~func_{\!\!,}\tilde ao}\\\underline{\rm dos~lados~e~do~\hat angulo}\\\underline{\rm compreendido~entre~eles}\\\huge\boxed{\boxed{\boxed{\boxed{\sf A=\dfrac{1}{2}\cdot a\cdot b\cdot sen(\theta)}}}}\\\sf a~e~b\longrightarrow lados~do~tri\hat angulo\\\sf \theta\longrightarrow \hat angulo~entre~os~lados\end{array}}$

$\Large\boxed{\begin{array}{l}\tt dados\begin{cases}\sf a=4~cm\\\sf b=5\sqrt{3}~cm\\\sf\theta=60^\circ\end{cases}\\\underline{\rm soluc_{\!\!,}\tilde ao:}\\\sf A=\dfrac{1}{2}\cdot a\cdot b\cdot sen(\theta)\\\sf A=\dfrac{1}{2}\cdot 4\cdot5\sqrt{3}\cdot sen(60^\circ)\\\sf A=\dfrac{1}{2}\cdot 4\cdot5\sqrt{3}\cdot\dfrac{\sqrt{3}}{2}\\\\\sf A=\dfrac{\diagdown\!\!\!\!\!20\cdot3}{\backslash\!\!\!4}\\\\\sf A=5\cdot3\\\sf A=15~cm^2\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~b}}}}\end{array}}$