Question

The sides of an acute triangle measure $14cm$, $18cm$, and $20cm$, respectively. Which of the following equations, when solved for $\theta$, gives the measure of the smallest angle of the triangle?
(Note: For nay triangle with sides of length $a$, $b$, and $c$ that are opposite angles $A$, $B$, and $C$, respectively, $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ and $c^2=a^2+b^2-2ab\cos C$).

• A. $\frac{\sin \theta }{14}=\frac{1}{18}$
• B. $\frac{\sin \theta }{14}=\frac{1}{20}$
• C. $\frac{\sin \theta }{20}=\frac{1}{14}$
• D. ${14}^2={18}^2+{20}^2-2\left(18\right)\left(20\right)\cos \theta$
• E. ${20}^2={14}^2+{18}^2-2\left(14\right)\left(18\right)\cos \theta$

Answer 1

The correct option is D ${14}^2={18}^2+{20}^2-2\left(18\right)\left(20\right)\cos \theta$

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

$\cos B=\frac{c^2+a^2-b^2}{2ca}$

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

Since, side with $14$ cm has the smallest opposite Angle

So, According to cosine Rule

$\cos \theta =\frac{{18}^2+{20}^o-{14}^2}{2\times 18\times 20}$

$\Rightarrow {14}^2={18}^2+{20}^2-2\left(18\right)\left(20\right)\cos \theta$