Question

In triangle $ABC,AB=5cm.\angle A={80}^o$ and $\angle B={70}^o$. Calculate the radius of the circumcircle and length of the other two sides.
(Necessary values can be taken from the following table)

Angle	sin	cos	tan
${70}^o$	0.94	0.34	2.75
${80}^o$	0.98	0.17	5.67

Answer 1

In $△ABC,$
$\angle A+\angle B+\angle C={180}^o$
${80}^o+{70}^o+\angle C={180}^o$
$\angle C={30}^o$
$\frac{BC}{\sin A}=\frac{AC}{\sin B}=\frac{AB}{\sin C}=diameter$
$\frac{AB}{\sin C}=diameter$
$\Rightarrow \frac{5}{\sin {30}^o}=diameter$
$\Rightarrow diameter=10cm$
$\Rightarrow radius=5cm$
$\frac{BC}{\sin A}=\frac{AB}{\sin C}$
$\Rightarrow \frac{BC}{\sin {80}^o}=\frac{5}{\sin {30}^o}$
$\Rightarrow BC=5times2\times 0.98$
$\Rightarrow BC=9.8cm$
$\frac{AC}{\sin B}=\frac{AB}{\sin C}$
$\Rightarrow \frac{AC}{\sin {70}^o}=\frac{5}{\sin {30}^o}$
$\Rightarrow AC=5times2\times 0.94$
$\Rightarrow AC=9.4cm$
Hence radius of the circle is $5cm$ and lengths of the other two sides are BC = 9.8cm and AC = 9.4cm.