In right triangle \( \mathrm{ABC}, \mathrm{m} \angle A=90^{\circ}, \sin B=2 x-0.28 \), and \( \cos C=4 x+0.16 \). Solve for \( x \).

To solve for \( x \) in the given right triangle \( \mathrm{ABC} \) with \( \angle A = 90^{\circ} \), \( \sin B = 2x - 0.28 \), and \( \cos C = 4x + 0.16 \), we can use the trigonometric identities for sine and cosine in a right-angled triangle. Given: - \( \angle A = 90^{\circ} \) - \( \sin B = 2x - 0.28 \) - \( \cos C = 4x + 0.16 \) We know that in a right-angled triangle, \( \sin B = \cos C \) because the sine of an angle is equal to the cosine of its complementary angle. Therefore, we can equate \( \sin B \) and \( \cos C \) to solve for \( x \): \[ 2x - 0.28 = 4x + 0.16 \] Now, we can solve this equation to find the value of \( x \). Solve the equation by following steps: - step0: Solve for \(x\): \(2x-0.28=4x+0.16\) - step1: Move the expression to the left side: \(2x-4x=0.16+0.28\) - step2: Add and subtract: \(-2x=0.16+0.28\) - step3: Add and subtract: \(-2x=0.44\) - step4: Change the signs: \(2x=-0.44\) - step5: Divide both sides: \(\frac{2x}{2}=\frac{-0.44}{2}\) - step6: Divide the numbers: \(x=-\frac{11}{50}\) The solution to the equation \(2x - 0.28 = 4x + 0.16\) is \(x = -\frac{11}{50}\) or \(x = -0.22\).