Question

In triangle ABC, ∠A = 35°, a = 18, and b = 24. (a) Show that there are two triangles, ABC and A1B1C1, that satisfy these conditions. Using the Law of Sines, we know the following. (Round your answers to three decimal places.) sin(B) ≈ ∠B ≈ ° and ∠B1 ≈ 180° − ° ≈ ° For triangle ABC, we see the following. (Round your answers to two decimal places.) ∠C ≈ ° and c ≈ For triangle A1B1C1, we see the following. (Round your answers to two decimal places.) ∠C1 ≈ ° and c ≈ 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 the following. Area of ΔABC / Area of ΔA1B1C1 = (1/2 ab sin(C)) / (1/2 ab sin(C1)) =

          In triangle ABC, ∠A = 35°, a = 18, and b = 24.
(a) Show that there are two triangles, ABC and A1B1C1, that satisfy these conditions.
Using the Law of Sines, we know the following. (Round your answers to three decimal places.)
sin(B) ≈ 
∠B ≈ ° and ∠B1 ≈ 180° − ° ≈ °

For triangle ABC, we see the following. (Round your answers to two decimal places.)
∠C ≈ °
and c ≈ 

For triangle A1B1C1, we see the following. (Round your answers to two decimal places.)
∠C1 ≈ °
and c ≈ 

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 the following.
Area of ΔABC / Area of ΔA1B1C1 = (1/2 ab sin(C)) / (1/2 ab sin(C1)) =

Added by Madison P.

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