Question

Find the area of the right angled triangle with hypotenuse 5 cm with one of the acute angle equal to 19°59'.
(Refer the trigonometric table for the values)

• A. 4.0155592 ${cm}^2$
• B. 4.0152955 ${cm}^2$
• C. 4.0152559 ${cm}^2$
• D. 4.0192555 ${cm}^2$

Answer 1

The correct option is B 4.0152955 ${cm}^2$

From the figure,
$sin\theta =\frac{AB}{AC}$
$sin19°{59}^{\prime }=\frac{AB}{5}$

0.3404 + 0.0014 $=\frac{AB}{5}$
[$\because$ sin19°59' = 0.3404 + 0.0014]

5 $\times$ 0.3418 = AB
AB = 1.709 cm

From the figure,
$cos\theta =\frac{BC}{AC}$
$cos19°{59}^{\prime }=\frac{BC}{5}$
0.9403 - 0.0005 $=\frac{BC}{5}$ [$\because$ cos19°59' = 0.9403 - 0.0005]
5$\times$ 0.9398 = BC
BC = 4.699 cm

$Areaofrighttriangle=\frac{1}{2}\times bh$

$=\frac{1}{2}\times BC\times AB$

$=\frac{1}{2}\times 4.699\times 1.709$
$=\frac{1}{2}\times 8.030591$
$=4.0152955{cm}^2$