HELP PLZZ will give brainliest <3Given the measures a = 10, b = 40, and A = 30°, how many triangles can possibly be formed?Given the measures b = 10, c = 8.9, and B = 63°, how many triangles can possibly be formed?

Answer:01Step-by-step explanation:First question:You are given a side, a, and its opposite angle, A. You are also given side b. Use that in the law of sines and solve for the other angle, B.[tex] \dfrac{a}{\sin A} = \dfrac{b}{\sin B} [/tex][tex] \dfrac{10}{\sin 30^\circ} = \dfrac{40}{\sin B} [/tex][tex] \dfrac{1}{0.5} = \dfrac{4}{\sin B} [/tex][tex] \sin B = 2 [/tex]The sine function can never equal 2, so there is no triangle in this case.Answer: no triangleSecond question:You are given a side, b, and its opposite angle, B. You are also given side c. Use that in the law of sines and solve for the other angle, C.[tex] \dfrac{b}{\sin B} = \dfrac{c}{\sin C} [/tex][tex] \dfrac{10}{\sin 63^\circ} = \dfrac{}{\sin C} [/tex][tex] \sin C = \dfrac{8.9\sin 63^\circ}{10} [/tex][tex] C = \sin^{-1} \dfrac{8.9\sin 63^\circ}{10} [/tex][tex] C \approx 52.5^\circ [/tex]One triangle exists for sure. Now we see if there is a second one.Now we look at the supplement of angle C.m<C = 52.5°supplement of angle C: m<C' = 180° - 52.5° = 127.5°We add the measures of angles B and the supplement of angle C:m<B + m<C' = 63° + 127.5° = 190.5°Since the sum of the measures of these two angles is already more than 180°, the supplement of angle C cannot be an angle of the triangle.Answer: one triangle