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a triangle abc with \\(\\angle c = 90^{circ}\\), (bc = 5\\text{ cm}), (…

Question

a triangle abc with \\(\angle c = 90^{circ}\\), (bc = 5\text{ cm}), (ab = 10\text{ cm}), and (\angle a=30^{circ}). options: (\tan(30^{circ})=\frac{5}{b}), (\tan(30^{circ})=\frac{b}{5}), (\tan(30^{circ})=\frac{10}{b}), (\tan(30^{circ})=\frac{b}{10})

Explanation:

Step1: Recall tangent - ratio definition

In a right - triangle, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$ with respect to an acute angle $\theta$. In right - triangle $ABC$ with right - angle at $C$, for angle $A = 30^{\circ}$, the side opposite to angle $A$ is $BC = 5$ cm and the side adjacent to angle $A$ is $AC=b$.

Step2: Apply the tangent formula

We know that $\tan(A)=\frac{\text{opposite}}{\text{adjacent}}$. Since $A = 30^{\circ}$, the opposite side to $\angle A$ is $BC = 5$ and the adjacent side is $AC = b$. So, $\tan(30^{\circ})=\frac{5}{b}$.

Answer:

$\tan(30^{\circ})=\frac{5}{b}$