Figure 1 shows the triangle ABC, with AB=6cm, BC=4cm and CA=5cm
a) show that cosA=3/4
b) hence, or otherwise, find the exact value of sinA
thx!
$\displaystyle c^2=a^2+b^2-2abcos(\theta)$ imput correct values and solve for $\displaystyle \theta$...and for the second one use the law of sines which says that $\displaystyle \frac{a}{sin(A)}=\frac{b}{sin(B)}=\frac{c}{sin(C)}$ where caps are angles and lower case are sides