Question

Решите треугольник abc,если1)b=12,угола=36 градусов,угол b=25градусов 2)b=18,c=8,угол а=130 градусов 3)а=10,b=12,c=8. !

Answer 1

1) $a=b \frac{sinA}{sinB} =12 \frac{sin36}{sin25}=16.69$ см
∠С=180°-(25°+36°)=119°
$c=b \frac{sin C}{sin A}=12 \frac{sin119}{sin36} =17.86$ см
2) a²=b²+b²-2bc·cosA
$a= \sqrt{ 18^{2}+ 8^{2} -2*18*8*cos130 }= \sqrt{573.12}=29.94$ см
≈$\frac{a}{sinA} = \frac{b}{sinB} \\ \\ sinB=b \frac{sinA}{a} =18 \frac{sin130}{23.94}=0,57$
∠B=35°10'4.15"
∠C=180°-130°-35°10'4.15"=14°49'55.9"
3) a²=b²+c²-2bc·cosB
$cosA= \frac{ b^{2} + c^{2} - a^{2} }{2bc} \\ \\ cosA= \frac{ 12^{2} + 8^{2}- 10^{2} }{2*12*8} =0.5625$
∠A=55°46'16.11"
b²=a²+c²-2ac·cosB
$cosB= \frac{ a^{2} + c^{2} - b^{2} }{2ac} = \frac{ 10^{2} + 8^{2}- 12^{2} }{2*10*8} =0.125$
∠B=82°49'9.28"
∠C=180°-55°46'16.1"-82°49'9.28"=41°24'34.6"