Question

Solve triangle ABC if ?A = 43.1°, a = 188.7, and b = 243.4. sin B = (round answer to 5 decimal places) There are two possible angles B between 0° and 180° with this value for sine. Find the two angles, and report them so that ?B1 is the acute angle. ?B1 = ° and ?B2 = ° (round these and all remaining answers to 1 decimal place) Thus, two triangles satisfy the given conditions: triangle A1B1C1 and triangle A2B2C2. Solve the first triangle: A1B1C1 ?C1 = ° and c1 = Solve the second triangle: A2B2C2 ?C2 = ° and c2 =

          Solve triangle ABC if ?A = 43.1°, a = 188.7, and b = 243.4.
sin B =
(round answer to 5 decimal places)
There are two possible angles B between 0° and 180° with this value for sine. Find the two angles, and report them so that ?B1 is the acute angle.
?B1 = ° and ?B2 = °
(round these and all remaining answers to 1 decimal place)
Thus, two triangles satisfy the given conditions: triangle A1B1C1 and triangle A2B2C2.
Solve the first triangle: A1B1C1
?C1 = ° and c1 =
Solve the second triangle: A2B2C2
?C2 = ° and c2 =

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Solve triangle ABC if ?A = 43.1°, a = 188.7, and b = 243.4.
sin B =
(round answer to 5 decimal places)
There are two possible angles B between 0° and 180° with this value for sine. Find the two angles, and report them so that ?B1 is the acute angle.
?B1 = ° and ?B2 = °
(round these and all remaining answers to 1 decimal place)
Thus, two triangles satisfy the given conditions: triangle A1B1C1 and triangle A2B2C2.
Solve the first triangle: A1B1C1
?C1 = ° and c1 =
Solve the second triangle: A2B2C2
?C2 = ° and c2 =

Added by Rich A.

Higher Engineering Mathematics

John Bird 8th Edition

Chapter 12

Instant Answer

Solved by Expert Aman Gupta

05/18/2023

Step 1

We can use the sine law for this: sin(B) = (b * sin(A)) / a sin(B) = (243.4 * sin(43.1°)) / 188.7 sin(B) ≈ 0.97497 (rounded to 5 decimal places) Now, we need to find the two possible angles B between 0° and 180° with this value for sine. We can use the inverse Show more…

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Solve triangle ABC if ∠A=43.1°, a=188.7, and b=243.4. There are two possible angles B between 0° and 180° with this value for sine. Find the two angles, and report them so that ∠B1 is the acute angle. ∠B1= ° and ∠B2= ° Thus, two triangles satisfy the given conditions: triangle A1B1C1 and triangle A2B2C2. Solve the first triangle: A1B1C1 ∠C1= ° and c1= Solve the second triangle: A2B2C2 ∠C2= ° and c2=

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