Can someone help me solve this problem?

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Answer 1 · 2018-03-05T00:58:43

The fourth answer down is correct:

#A=1/2bh=1/2efsinD#

.

The small triangle to the left of the altitude #h# is a right angle triangle. We can write:

#sinD=("Opposite")/("Hypotenuse")=h/f#

We can solve for #h#:

#h=fsinD#

In the formula for the Area of the triangle #Delta DEF#, we can plug in #e# for base #b#, and #fsinD# for #h#:

#A=1/2bh=1/2efsinD#