It is given that in △abc, ad ⊥ bc. using the definition of sine with angle c in △acd results in sin(c) = . using the multiplication property of equality to isolate h, the equation becomes bsin(c) = h. knowing that the formula for the area of a triangle is a = bh is and using the side lengths as shown in the diagram, which expression represents the area of △abc? bsin(c) absin(c) cbsin(c) hbsin(c)

We see here in the diagram that the base is a. We know this because the height is perpendicular to it. We also know the height is bsin(C) which, when replace h for bsin(C) and a for the base, we get A=absin(C), which is the second option.

B

Step-by-step explanation:

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

answer: i'm great at algebra

~coco

step-by-step explanation: