Question

4. 5. 6. 7. нужно срочно ​

Answer 1

4) 1

$\frac{AB}{sin(C)} =\frac{BC}{sin(A)} \\BC=\frac{AB*sin(30)}{sin(45)} =\frac{\sqrt{2} *\frac{1}{2} }{\frac{\sqrt{2} }{2} } =\frac{\sqrt{2} *2}{\sqrt{2} *2} =1$

5) 5

$a^{2} =b^{2} +c^{2} -2bc*cosA\\a=\sqrt{b^{2} +c^{2} -2bc*cosA} =\sqrt{4^{2} +5^{2}-2*4*5*0,4 } =\sqrt{16+25-16} =\sqrt{25}=5$

6) 15/17

$\frac{KL}{sin(M)} =\frac{LM}{sin(K)} \\sin(K)=\frac{sin(M)*LM}{KL} =\frac{\frac{8}{17} *17}{8} =\frac{8*17}{17*8} =1$

<K = 90° , значит треугольник прямоугольный

$KM = \sqrt{LM^{2} -KL^{2} } =\sqrt{17^{2}-8^{2} } =\sqrt{(17-8)(17+8)} =\sqrt{9*25} =3*5=15$

$sin(L)=\frac{KM}{LM} =\frac{15}{17}$

7) 0,5

$\frac{KL}{sin(M)} =\frac{LM}{sin(K)}\\\frac{2LM}{sin(M)} =\frac{LM}{sin(K)}\\sin(M)=\frac{2LM*sin(K)}{LM} =2sin(K)=2*\frac{1}{4} =\frac{1}{2} =0,5$