Question

Answer 1

Lời giải:

Áp dụng định lý Pitago:

$BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+15^2}=3\sqrt{29}$

$\sin B = \frac{AC}{BC}=\frac{15}{3\sqrt{29}}=\frac{5}{\sqrt{29}}$

$\cos B=\frac{AB}{BC}=\frac{6}{3\sqrt{29}}=\frac{2}{\sqrt{29}}$

$\tan B=\frac{AC}{AB}=\frac{15}{6}=\frac{5}{2}$

$\cot B= \frac{AB}{AC}=\frac{2}{5}$

$\sin C=\cos B=\frac{2}{\sqrt{29}}$

$\cos C=\sin B =\frac{5}{\sqrt{29}}$

$\tan C=\cot B=\frac{2}{5}$

$\cot C=\tan B = \frac{5}{2}$

Answer 2

tam giác ABC vuông tại A có
\(BC^2=AB^2+AC^2\)
\(BC^2=6^2+15^2=261\)
BC=\(3\sqrt{29}\)cm
\(sinB=\dfrac{5\sqrt{29}}{29};cosB=\dfrac{2\sqrt{29}}{29};tanB=\dfrac{5}{2};cotanB=\dfrac{2}{5}\)
\(cosc=\dfrac{5\sqrt{29}}{29};sinC=\dfrac{2\sqrt{29}}{29};cotanB=\dfrac{5}{2};tanB=\dfrac{2}{5}\)