Question

I Given A=190, 6:12, and a 10, use the law or sines to solve The triangle (if possible) for the value of c. If two Solutions exist, Riad both. Round ansuec to two decimal places.

Answer 1

ANSWER ::

Possible solutions ::

1) B = 23 Degrees , C = 138 Degrees , c = 20.55

2) B = 157 Degrees , C = 4 Degrees , c = 2.14

Given A - 199 a = 10 b = 12 с Possible Solution - I b=12 as 10 ) 190 From Law of Sines, A с SinB Sin A' elo 12 (Simr)] Sin B 12 B arcsin Sin 19° 10 22.997 B 23 (rounded to two decimal) From triangle Properties, A+B+= 180° 19+ 25 + c = 180 c=180-(19+23) 1389 = Now, for Side ;- Using law of Sines, Sinc Sin A 10 Sin 138 Sin 19 Sin138 Sin 19° (10) ANSWER:- B = 23 с 20.55 c=138 C = 20.55 CS Scanned with CamScanner

Possible Solution I b=12 a10 199 From the Law of Sines, A Sin B SinA a SinB bo (Sink) iso-arcsin ( (Sina")) BE 180 - arc Sivil (Sin 19)) 157-003 B = 157" Now, from triangle Properties, A+ 6+ c = 180° 19.+ 157°+c=180 C = 180-(191157) c = 3.997° Ć - 4 Now, for Side C ; - sines, Using Law of Sinc SinA sin 4 Sinige 10 C = Sin 4 (10) Sin 19" c = 2.14 ANSWER; B = 157 c = 4° O C = 2.14 CS Scanned with CamScanner