(a) Show that there are two triangles, ABC and
A1B1C1,that satisfy these conditions.
Using the Law of Sines, we know that (Round your answers to three decimal places.)
sin B ≈ .857
B ≈ 59.987
° and B1 ≈ 180° − 58.987
° ≈ 121.013 °
For triangle ABC, we see that (Round your answers to two decimal places.)
C ≈ 81.013 °
and c ≈ 23.049
For triangle
A1B1C1,
we see that (Round your answers to two decimal places.)
C1 ≈ 18.987 °
and c ≈ 7.592
Thus, there are two triangles that satisfy these conditions.
(b) Show that the areas of the triangles in part (a) are proportional to the sines of the angles C and C1, that is,
area of ΔABC/
area of ΔA1B1C1 =
sin C/
sin C1
.
By the area formula, we see that
Area of ΔABC/
Area of ΔA1B1C1 =
1/2 ab sin C/1/2 ab sin C1 =
I don't know what the final answer is? Or what it's asking for.... Please help.
Answers & Comments
Verified answer
If we look at AC ( = "b" ) as the "base" of the triangle,
then the area is (1/2) b times the "altitude".
But the "altitude" is 15 sin C or 15 sin C1.
So one area is 150 sin C, and the other is 150 sin C1.
That's all they're really asking you to recognize,
that the ratio of the areas is the same as the ratio
(sin C)/(sin C1).
It's very possible that what you wrote down was an
attempt to say exactly the same thing I've just said!