(8) In triangle \( A B C \) the angle at \( B \) is \( 90^{\circ} \). If \( \mathrm{BC}=23 \mathrm{~mm} \) and \( \mathrm{AB}=37 \mathrm{~mm} \) find the size of the angle at C in degrees. angle \( \mathrm{C}= \)
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To find the size of angle \( C \) in triangle \( ABC \), we can use the sine or cosine function. In this case, we can apply the cosine rule, which states: \[ \cos(C) = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC} \] First, we need to find the length of side \( AC \) using the Pythagorean theorem since \( B \) is \( 90^\circ \): \[ AC = \sqrt{AB^2 + BC^2} = \sqrt{37^2 + 23^2} = \sqrt{1369 + 529} = \sqrt{1898} \approx 43.6 \text{ mm} \] Now we can substitute \( AB \), \( BC \), and \( AC \) into the cosine formula: \[ \cos(C) = \frac{37^2 + 23^2 - (43.6)^2}{2 \cdot 37 \cdot 23} \] Calculating this gives us: \[ \cos(C) = \frac{1369 + 529 - 1898}{2 \cdot 37 \cdot 23} = \frac{0}{1702} = 0 \] Since \( \cos(C) = 0 \), angle \( C = 90^\circ \). Hence, angle \( C = 90^\circ \) in triangle \( ABC \).
