Respuesta :
Answer:
0
1
Step-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 triangle
Second 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
Answer:
The answer to the first one is 0
The answer to the second one is 1
Step-by-step explanation:
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