In the triangle ABC, a = 7, c = 5 and A = 68°. Find the perimeter of the triangle.
Find the perimeter of the triangle ABC if a = 7.8, b = 6.2 and A = 50°
can someone please help me with these 2 questions and can you als provide working out. Thanks
+33661 Assuming A is the angle opposite side a, then we can use the cosine rule to get side b:
a2 = b2 + c2 - 2*b*c*cos(A)
72 = b2 + 52 - 2*b*5*cos(68°)
Rearrange to get the quadratic equation:
b2 - 10*cos(68°)*b - 24 = 0
$${{\mathtt{b}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{68}}^\circ\right)}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,-\,}}{\mathtt{24}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{b}} = {\mathtt{5}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{2\pi}}}{{cos}}{\left({\frac{{\mathtt{17}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{45}}}}\right)}{\mathtt{\,-\,}}{\sqrt{{\mathtt{25}}{\mathtt{\,\times\,}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{2\pi}}}{{cos}}{\left({\frac{{\mathtt{17}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{45}}}}\right)}}^{\,{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{24}}}}\\
{\mathtt{b}} = {\sqrt{{\mathtt{25}}{\mathtt{\,\times\,}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{2\pi}}}{{cos}}{\left({\frac{{\mathtt{17}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{45}}}}\right)}}^{\,{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{24}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{2\pi}}}{{cos}}{\left({\frac{{\mathtt{17}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{45}}}}\right)}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{b}} = -{\mathtt{3.371\: \!798\: \!059\: \!346\: \!734}}\\
{\mathtt{b}} = {\mathtt{7.117\: \!863\: \!993\: \!506\: \!734}}\\
\end{array} \right\}$$
There is only one positive solution for b, so take this and add it to a and c to find the perimeter.
Use the same technique on part 2.
.
+33661 Assuming A is the angle opposite side a, then we can use the cosine rule to get side b:
a2 = b2 + c2 - 2*b*c*cos(A)
72 = b2 + 52 - 2*b*5*cos(68°)
Rearrange to get the quadratic equation:
b2 - 10*cos(68°)*b - 24 = 0
$${{\mathtt{b}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{68}}^\circ\right)}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,-\,}}{\mathtt{24}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{b}} = {\mathtt{5}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{2\pi}}}{{cos}}{\left({\frac{{\mathtt{17}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{45}}}}\right)}{\mathtt{\,-\,}}{\sqrt{{\mathtt{25}}{\mathtt{\,\times\,}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{2\pi}}}{{cos}}{\left({\frac{{\mathtt{17}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{45}}}}\right)}}^{\,{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{24}}}}\\
{\mathtt{b}} = {\sqrt{{\mathtt{25}}{\mathtt{\,\times\,}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{2\pi}}}{{cos}}{\left({\frac{{\mathtt{17}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{45}}}}\right)}}^{\,{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{24}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{2\pi}}}{{cos}}{\left({\frac{{\mathtt{17}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{45}}}}\right)}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{b}} = -{\mathtt{3.371\: \!798\: \!059\: \!346\: \!734}}\\
{\mathtt{b}} = {\mathtt{7.117\: \!863\: \!993\: \!506\: \!734}}\\
\end{array} \right\}$$
There is only one positive solution for b, so take this and add it to a and c to find the perimeter.
Use the same technique on part 2.
.
