Question

What is the area of the triangle below? Use sin 40 = 0.64.

slanted side: 22, bottom:29, inside corner: 40

A. 200.11

B. 202.61

C. 204.16

D. 224.61

Answer 1

We have that angle A =40 deg, AB=22, AC=29

we know that area $= 0.5^{*}\mathrm{AB}^{*}\mathrm{AC}^{*}\sin (\mathrm{angle~A})$

$\sin (\mathrm{A}) = \sin (40) = 0.64$

then area $= 0.5^{*}22^{*}29^{*}0.64 = 204.16$

Correct answer C.