Question

Find the area of a right triangle whose hypotenuse is 10 cm and one of the acute angle is 24°24′. (Given: sin 24° 24′ = 0.4131, cos 24° 24′ = 0.9107)

• A. $15c{m}^2$
• B. $16.59c{m}^2$
• C. $17.57c{m}^2$
• D. $18.81c{m}^2$

Answer 1

The correct option is D $18.81c{m}^2$

In the $\Delta ABC,$
sin $24°24\text{'}=\frac{AB}{AC}$
$0.4131=\frac{AB}{10}$
$\therefore AB=4.131cm$

In the $\Delta ABC,$
cos $24°24\text{'}=\frac{BC}{AC}$
$0.9107=\frac{BC}{10}$
$\therefore BC=9.107$ cm

Area of the right angle
$=\frac{1}{2}\times BC\times AB$
$=\frac{1}{2}\times 9.107\times 4.131$
$=\frac{37.62}{2}$
$=18.81$

Area of the triangle $=18.81cm²$