Find all solutions to the following triangle. (Round your answers for angles A, C, A', and C' to the nearest whole number. Round your answers for sides c and c' to two decimal places. If either triangle is not possible, enter NONE in each corresponding answer blank.) B = 117°, b = 0.66 cm, a = 0.96 cm First triangle (assume A ≤ 90°): A = ° C = ° c = cm Second triangle (assume A' > 90°): A' = ° C' = ° c' = cm
Find all solutions to the following triangle. (Round your answers for angles A, C, A', and C' to the nearest whole number. Round your answers for sides c and c' to two decimal places. If either triangle is not possible, enter NONE in each corresponding answer blank.) B = 117°, b = 0.66 cm, a = 0.96 cm First triangle (assume A ≤ 90°): A = ° C = ° c = cm Second triangle (assume A' > 90°): A' = ° C' = ° c' = cm
Find all solutions to the following triangle. (Round your answers for angles A, C, A', and C' to the nearest whole number. Round your answers for sides c and c' to two decimal places. If either triangle is not possible, enter NONE in each corresponding answer blank.) B = 117°, b = 0.66 cm, a = 0.96 cm First triangle (assume A ≤ 90°): A = ° C = ° c = cm Second triangle (assume A' > 90°): A' = ° C' = ° c' = cm
Find all solutions to the following triangle. (Round your answers for anglesA, C, A', and C' to the nearest whole number. Round your answers for sides c and c' to two decimal places. If either triangle is not possible, enter NONE in each corresponding answer blank.)
B = 117°, b = 0.66 cm, a = 0.96 cm
First triangle (assume A ≤ 90°):
A
=
°
C
=
°
c
=
cm
Second triangle (assume A' > 90°):
A'
=
°
C'
=
°
c'
=
cm
Figure in plane geometry formed by two rays or lines that share a common endpoint, called the vertex. The angle is measured in degrees using a protractor. The different types of angles are acute, obtuse, right, straight, and reflex.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, trigonometry and related others by exploring similar questions and additional content below.
