Question

Trigonometry, how would I find the area of this triangle?

Answer 1

Use sine rule for Area

$\\ \rm\rightarrowtail Area=\dfrac{1}{2}absinC$

$\\ \rm\rightarrowtail Area=\dfrac{1}{2}(3.9)(5.6)sin56°45'$

$\\ \rm\rightarrowtail Area=10.92(0.836)$

$\\ \rm\rightarrowtail Area=10.22cm^2$

Answer 2

Answer:

Step-by-step explanation:

Area of triangle = 1/2 * base * height

Use the side of length 5.6cm as base, the height will be a perpendicular line from the base to the tip of the triangle.

From the right-angle triangle formed by the height and the side of length 3.9cm,

sin(56°45') = height / 3.9

height = 3.9*sin(56°45')

Substituting back into the area equation:

Area = 1/2*5.6*3.9*sin(56°45')

= 9.0cm^2

= 9cm^2