Question

solve for x to the nearest tenth.
triangle abc with right angle at c, bc = 6, angle at a is 75 degrees, ab = x

Explanation:

Step1: Identify triangle type and trigonometric ratio

This is a right - triangle \( \triangle ABC \) with \( \angle C = 90^{\circ} \), \( BC = 6 \), \( \angle A=75^{\circ} \), and we need to find \( AB=x \). We can use the cosine function. In a right - triangle, \( \cos A=\frac{AC}{AB} \), but we can also use \( \cos(90^{\circ}-\angle A)=\cos(15^{\circ})=\frac{BC}{AB} \) (since \( \angle B = 90^{\circ}-\angle A=15^{\circ} \)) or \( \sin\angle B=\frac{AC}{AB} \), but more directly, we know that \( \cos\angle A=\frac{AC}{AB} \) and \( \sin\angle A=\frac{BC}{AB} \). Wait, \( \sin\angle A=\frac{BC}{AB} \), so \( \sin75^{\circ}=\frac{6}{x} \).

Step2: Solve for x

From \( \sin75^{\circ}=\frac{6}{x} \), we can re - arrange the formula to get \( x = \frac{6}{\sin75^{\circ}} \). We know that \( \sin75^{\circ}=\sin(45^{\circ} + 30^{\circ})=\sin45^{\circ}\cos30^{\circ}+\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.9659 \)

Then \( x=\frac{6}{0.9659}\approx6.2 \)

Answer:

\( 6.2 \)