In a △ A B C , A B × A C = 4 8 and A D is the perpendicular dropped from A on B C ,such that A D = 6 .What should replace x in the below equation.
B C = sin ∠ A × x
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Good approach.
You seem to have made an implicit assumption that D is on the line segment BC. Must this be true? If not, how would the proof change?
Note: This is one reason why in geometry proofs, we make use of directed angles and directed line segments, to avoid such implicit assumptions.
It is given in the question that A D is a perpendicular dropped from A on B C .
I created this question using the following equation:
A B ∗ A C = 2 R ∗ A D ( R is the circumradius of △ A B C )
putting the values in the equation we get,
2 R = 8
Now, using the sine rule
sin ∠ A B C = 2 R
B C = 2 R ∗ sin ∠ A
x = 8
This problem could be solved using the formulae as A= 2 1 x B x H and A=1/2x pro. Of opp.sides x sin(a)
Since altitude is AD=6, AB, or AC minimum can be only 6. So let AB=6. So B and D coincides. AC=48/6=8. Angle B= angle D=90. B C = A C S i n B A C = A C B C B y S i n L a w , S i n B A C B C = A C B C B C = 8 .
This solution makes an assumption and deals with a specific case, instead of the general case.
Problem Loading...
Note Loading...
Set Loading...
sin ∠ A = sin ( x + y ) ⇒ sin ∠ A × 8 = sin x cos y + cos x sin y = A B B D ⋅ A C A D + A B A D ⋅ A C C D = 8 B D + 8 C D = 8 B C = B C