In questions 1-11 \(\Delta \mathrm{ABC}\) has a right angle at \(\mathrm{C}\). Calculate \(\mathrm{AC}\) given \(\mathrm{BC}=12 \mathrm{~cm}\) and \(B=53^{\circ}\)

We have a right triangle with a right angle at C. We are given the length of side BC as \(12 \mathrm{~cm}\) and angle B as \(53^{\circ}\). Our goal is to find the length of side AC. Let AC be denoted as 'a'.

In a right triangle, the sine of an angle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. In our triangle, sin(B) would be the ratio of the side opposite angle B (AC or 'a') to the hypotenuse (AB). We can write this as: \(\sin{B}=\frac{a}{\mathrm{AB}}\)

We want to find the length of side AC, or 'a'. We were given the value for angle B and the length of side BC. We can use the sine function to find 'a': \(\sin{53^{\circ}}=\frac{a}{\mathrm{AB}}\) Since \(\sin{53^{\circ}}\) is a known value, we can replace it with its decimal equivalent (approximately 0.8): \(0.8=\frac{a}{\mathrm{AB}}\) We can solve for 'a' by multiplying both sides by AB: \(a=0.8 \times \mathrm{AB}\) The relationship between the sides of a right triangle is given by the Pythagorean theorem: \(a^{2}+\mathrm{BC}^2=\mathrm{AB}^2\) We were given the length of side BC (\(12\,\mathrm{cm}\)), so we can plug that into the equation: \(a^{2}+144=\mathrm{AB}^2\) Since we found the value of 'a' in terms of AB in an earlier step, we can plug that expression into the equation: \((0.8 \times \mathrm{AB})^{2}+144=\mathrm{AB}^2\) Solve for AB: \((0.64)\mathrm{AB}^{2}+144=\mathrm{AB}^2\) \(144 = (1 - 0.64) \mathrm{AB}^2\) \(\frac{144}{0.36} = \mathrm{AB}^2\) \(\mathrm{AB} = \sqrt{400}=20\) Now that we have the value of AB, we can plug that back into the equation we found earlier for 'a': \(a=0.8 \times 20\) \(a=16\) Therefore, the length of side AC is \(16 \mathrm{~cm}\).