Consider any triangle (see figure). Define #sin theta = h/c#, show that the area of triangle is #A_Delta = 1/2 (b*c) sintheta# where #b and c# are any two sides of the traingle that make the angle #theta#?

Question

Evelyn Arboleda · Accepted Answer

See below.

It is well know that #A_Delta = 1/2 b h = 1/2 b h c/c = 1/2b c (h/c) = 1/2b c sin theta# the same regarding angle #hat(ACB)# #A_Delta = 1/2 b h = 1/2 b h a/a = 1/2b a (h/a) = 1/2b a sin beta# Also can be established # 1/2b c sin theta = 1/2b a sin beta->sin theta/a=sin beta/c#

Parker Allen · Answer

undefined To show that the area $ A_{\Delta} $ of a triangle can be expressed as $ \frac{1}{2}bc\sin\theta $, where $ b $ and $ c $ are any two sides of the triangle making an angle $ \theta $, we can use the formula for the area of a triangle given the length of one side and the height corresponding to that side.

1. Consider a triangle with sides $ b $, $ c $, and height $ h $ corresponding to side $ c $.
2. By definition, $ \sin\theta = \frac{h}{c} $, where $ \theta $ is the angle opposite side $ c $.
3. We know that the area $ A_{\Delta} $ of a triangle can be expressed as $ \frac{1}{2} \times \text{base} \times \text{height} $.
4. Substituting $ b $ for the base and $ h $ for the height, we get $ A_{\Delta} = \frac{1}{2}bh $.
5. Substitute $ h = c\sin\theta $ into the formula, we get $ A_{\Delta} = \frac{1}{2}b(c\sin\theta) $.
6. Simplify to obtain $ A_{\Delta} = \frac{1}{2}bc\sin\theta $.
   
Thus, we have shown that the area of the triangle can indeed be expressed as $ \frac{1}{2}bc\sin\theta $, where $ b $ and $ c $ are any two sides of the triangle making an angle $ \theta $.