Trigonometry

[Guest]

If side P = 8cm, Q = 7cm and side R = 5cm. So the area of the triangle is?

[Guest]

a=8 ; b=7 ; c=5 ; [tex]S_{ABC }[/tex]=?

(cos T) [tex]a^{2}[/tex]=[tex]b^{2}[/tex]+[tex]c^{2}[/tex]-2bc.cos[tex]\alpha[/tex]

cos[tex]\alpha[/tex]=[tex]\frac{b^{2}+c^{2}-a^{2}}{2bc}[/tex]

cos[tex]\alpha[/tex]=[tex]\frac{49+25-64}{2.7.5}[/tex]

cos[tex]\alpha[/tex]=[tex]\frac{1}{7}[/tex] [tex]\Leftrightarrow[/tex] sin[tex]\alpha[/tex]=[tex]\sqrt{\frac{48}{49}}[/tex]

sin[tex]\alpha[/tex]=[tex]\frac{4\sqrt{3}}{7}[/tex]

[tex]S_{ABC }[/tex]=[tex]\frac{bc.sin\alpha}{2}[/tex]=[tex]\frac{7.5.4\sqrt{3}}{2.7}[/tex]

[tex]S_{ABC }[/tex]=10[tex]\sqrt{3}[/tex]

[Baltuilhe]

Good evening!

Another way to solve the same problem !

Using Heron's Formula.
[tex]p=\dfrac{a+b+c}{2}=\dfrac{8+7+5}{2}=\dfrac{20}{2}=10\\
A=\sqrt{p\cdot\left(p-a\right)\cdot\left(p-b\right)\cdot\left(p-c\right)}\\
A=\sqrt{10\cdot\left(10-8\right)\cdot\left(10-7\right)\cdot\left(10-5\right)}\\
A=\sqrt{10\cdot 2\cdot 3\cdot 5}\\
A=\sqrt{300}\\
\boxed{A=10\sqrt{3}}[/tex]

I hope I have helped!