Trianlge A B C has A B = 33 , A C = 88 , B C =

rzfansubs87

rzfansubs87

Answered question

2022-07-05

Trianlge A B C has A B = 33 , A C = 88 , B C = 77. Point D lies on B C with B D = 21. Compute B A D .
By the angle bisector theorem we have 21 56 = 33 88 and we know that A D bisects B A C. Now from here I was able to use the law of cosines, but it resulted in a pretty ugly looking expression
77 2 = 33 2 + 88 2 2 ( 33 ) ( 88 ) c o s ( B A C )
from here I was pretty much forced to use a calculator to figure out that
B A C = 60 B A D = 30.
After looking at the solution they had a very similar approach, but instead of
77 2 = 33 2 + 88 2 2 ( 33 ) ( 88 ) c o s ( B A C )
they had
7 2 = 3 2 + 8 2 2 ( 3 ) ( 8 ) c o s ( B A C )
which I don’t really get. How can you take one digit of from every term there?

Answer & Explanation

Oliver Shepherd

Oliver Shepherd

Beginner2022-07-06Added 24 answers

This isn't a question of geometry.....its simple arithmetic:
77 2 = 33 2 + 88 2 2 ( 33 ) ( 88 ) c o s ( B A C )
Here you can see every term has a factor of 11 2 .
( 11 2 ) 7 2 = 3 2 ( 11 ) 2 + 8 2 ( 11 ) 2 ( 11 2 ) 2 ( 3 ) ( 8 ) c o s ( B A C )
You can cancel out this factor by dividing by 11 2 on both sides.
7 2 = 3 2 + 8 2 2 ( 3 ) ( 8 ) c o s ( B A C )

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