a)
Tam giác ABC vuông tại A nên ta có: \(B{C^2} = A{B^2} + A{C^2}\) (Định lý Pythagore)
Thay số ta có \({17^2} = {8^2} + A{C^2}\) hay \(A{C^2} = {17^2} - {8^2} = 225\) suy ra \(AC = 15\) cm (vì \(AC > 0\))
Ta có: \(\sin \widehat B = \cos \widehat C = \frac{{AC}}{{BC}} = \frac{{15}}{{17}}\)
\(\cos \widehat B = \sin \widehat C = \frac{{AB}}{{BC}} = \frac{8}{{17}}\)
\(\tan \widehat B = \cot \widehat C = \frac{{AC}}{{AB}} = \frac{{15}}{8}\)
\(\cot \widehat B = \tan \widehat C = \frac{{AB}}{{AC}} = \frac{8}{{15}}\)
b)
Tam giác ABC vuông tại A nên ta có: \(B{C^2} = A{B^2} + A{C^2}\) (Định lý Pythagore)
Thay số ta có \(B{C^2} = 1,{2^2} + 0,{9^2} = 2,25\) hay \(CB = \sqrt {2,25} = 1,5\) cm (vì \(BC > 0\))
Ta có: \(\sin \widehat B = \cos \widehat C = \frac{{AC}}{{BC}} = \frac{{0,9}}{{1,5}} = \frac{3}{5}\)
\(\cos \widehat B = \sin \widehat C = \frac{{AB}}{{BC}} = \frac{{1,2}}{{1,5}} = \frac{4}{5}\)
\(\tan \widehat B = \cot \widehat C = \frac{{AC}}{{AB}} = \frac{{0,9}}{{1,2}} = \frac{3}{4}\)
\(\cot \widehat B = \tan \widehat C = \frac{{AB}}{{AC}} = \frac{{1,2}}{{0,9}} = \frac{4}{3}\)