Question

question 4(multiple choice worth 1 points) (05 02 mc) in triangle jkl, sin(b°) = 3/5 and cos(b°) = 4/5. if triangle jkl is dilated by a scale factor of 2, what is tan(b°)?

Explanation:

Step1: Recall tangent - sine - cosine relationship

In a right - triangle, $\tan\theta=\frac{\sin\theta}{\cos\theta}$.

Step2: Substitute the given values

Given $\sin(b^{\circ})=\frac{3}{5}$ and $\cos(b^{\circ})=\frac{4}{5}$. Then $\tan(b^{\circ})=\frac{\sin(b^{\circ})}{\cos(b^{\circ})}=\frac{\frac{3}{5}}{\frac{4}{5}}$.

Step3: Simplify the fraction

$\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{5}\times\frac{5}{4}=\frac{3}{4}$. Dilation does not change the angles and their trigonometric ratios.

Answer:

$\frac{3}{4}$