Question

Find the area of the right triangle given in figure

Answer 1

From the figure, $\sin \theta =\frac{AB}{AC}\Rightarrow \sin {10}^{\circ }{14}^{\prime }=\frac{AB}{3}$
From the sine table, $\sin {10}^{\circ }{12}^{\prime }=0.1771$ and Mean Difference for $2^{\prime }=0.0006$
$\therefore \sin {10}^{\circ }{14}^{\prime }=0.1777\Rightarrow 0.1777=\frac{AB}{3}$
$\therefore AB=0.1777\times 3=0.5331cm$
$\cos \theta =\frac{BC}{AC}\Rightarrow \cos {10}^{\circ }{14}^{\prime }=\frac{BC}{3}$
From the cosine table, $\cos {10}^{\circ }{12}^{\prime }=0.9842$ and Mean Difference for $2^{\prime }=0.0001$
$\therefore \cos {10}^{\circ }{14}^{\prime }=0.9841-0.0001=0.9841$
$0.9841=\frac{BC}{3}$
$\therefore BC=0.9841\times 3=2.9523cm$
Area of the right triangle $=\frac{1}{2}bh=\frac{1}{2}\times 2.9523\times 0.5331$
$=0.786935565$
$\therefore$ Area of the triangle is $0.7869c{m}^2$ (approximately)