# Trigonometry

Trigonometry equalities, inequalities and expressions - sin, cos, tan, cot

### Trigonometry

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

### Re: Trigonometry

a=8 ; b=7 ; c=5 ; $$S_{ABC }$$=?

(cos T) $$a^{2}$$=$$b^{2}$$+$$c^{2}$$-2bc.cos$$\alpha$$

cos$$\alpha$$=$$\frac{b^{2}+c^{2}-a^{2}}{2bc}$$

cos$$\alpha$$=$$\frac{49+25-64}{2.7.5}$$

cos$$\alpha$$=$$\frac{1}{7}$$ $$\Leftrightarrow$$ sin$$\alpha$$=$$\sqrt{\frac{48}{49}}$$

sin$$\alpha$$=$$\frac{4\sqrt{3}}{7}$$

$$S_{ABC }$$=$$\frac{bc.sin\alpha}{2}$$=$$\frac{7.5.4\sqrt{3}}{2.7}$$

$$S_{ABC }$$=10$$\sqrt{3}$$
Guest

### Re: Trigonometry

Good evening!

Another way to solve the same problem !

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

I hope I have helped!

Baltuilhe

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