Question

emplea los teoremas del seno y del coseno para resolver los siguientes triángulos ABC:
1. a: 10 cm, b: 12 cm, c: 35° 40
2. A: 70° , B: 40° , c: 10cm
3. a: 10cm, b: 15cm, B:42°
4. A:52° 30’. B:78° 12’. c: 300,5 m
5. b: 4 km, a: 13km , A:53° 8

Answer 1

A continuación se muestran los datos faltantes para completar cada triángulo

1. c = 7.001, A = 56º 23' 28.23'', B = 88º 1' 15.32''
2. C = 70, a = 10, b = 6.84.
3. A = 41º 59', C = 96', c = 22.29cm
4. C = 49º18', a = 311.45m, b =387.99m
5. B = 14º14', C = 112º 37', c = 15km

Una vez sabido estos resultados, vamos a llegar a los mismos de una manera sencilla y explicada paso a paso

1. $c = \sqrt{a^{2} + b^{2} - 2abcos(\theta)} = \sqrt{10^{2} + 12^{2} - 2*120cos(35º40)} \\= \sqrt{100+144-240*0.812}  = \sqrt{49.018}= 7.001$, $\frac{sinA}{a} = \frac{sinC}{c} \\sin(A) = \frac{asinC}{c} = \frac{10sin(35º40'}{7.001} = 0.8328\\ A = sin^{-1}(0.8328) = 56º23'\\\frac{sinB}{b} = \frac{sinC}{c} \\sin(B) = \frac{bsinC}{c} = \frac{12sin(35º40'}{7.001} = 0.999\\ B = sin^{-1}(0.999) = 87º26'$
2. $C = 180 - A - B = 180-70-40 = 70\\\frac{a}{sinA} = \frac{c}{sinC}\\a = \frac{csin(A)}{sin(C)} = \frac{10sin(70)}{sin(70)} = 10\\\frac{b}{sinB} = \frac{c}{sinC}\\b = \frac{csin(B)}{sin(C)} = \frac{10sin(40)}{sin(70)} = 6.84$
3. $\frac{sinA}{a} = \frac{sinB}{b} \\sinA = \frac{asin(B)}{b} = \frac{10sin(42)}{15} = 0.669\\A = sin^{-1}(0.669)= 41º 59'\\C = 180-42 - (41º 59') = 96\\c = \frac{bsin(C)}{sin(B)} = \frac{15sin(96)}{sin(42)} = 22.29$
4. $C = 180-A-B = 49º 18'\\a = \frac{csin(A)}{sin(C)} = \frac{300.5sin(52º30')}{sin(49º18')} = 311.45m .\\b = \frac{csin(B)}{sin(C)} = \frac{300.5sin(78º12')}{sin(49º18')} = 387.99m .$
5. $sin(B)=\frac{bsin(A)}{a} = \fac{4sin(53º8')}{13} = 0.246\\B= sin^{-1}(0.246)=14º14'\\C = 180-A-B = 112º37'\\c = \sqrt{a^{2}+ b^{2}-2abcos(C)}\\ = \sqrt{13^{2}+ 4^{2}-104cos(112º37')}\\= \sqrt{185+39.99} = \sqrt{224.99}=14.999 km$