Respuesta :
we won´t have any triangle with these measures.
law of cosines.
a²=b²+c²-2bc(cosФ)
Data:
a=1
b=2
A=31º
Therefore:
1²=2²+c²-4c(cos 31º)
c²-(4cos 31º)c+3=0
4cos 31º≈3.43
We have to solve this equation, and find out the number of solutions:
c=[4cos 31º⁺₋√(11.756-12)]/2=
c=(4cos 31º⁺₋√(-0.244)/2
Because we have the square root of a negative number, we don´t have any possible solutions, that is to say , there are not real solutions for this equation.
Answer: We don´t have any triangle with these measures.
law of cosines.
a²=b²+c²-2bc(cosФ)
Data:
a=1
b=2
A=31º
Therefore:
1²=2²+c²-4c(cos 31º)
c²-(4cos 31º)c+3=0
4cos 31º≈3.43
We have to solve this equation, and find out the number of solutions:
c=[4cos 31º⁺₋√(11.756-12)]/2=
c=(4cos 31º⁺₋√(-0.244)/2
Because we have the square root of a negative number, we don´t have any possible solutions, that is to say , there are not real solutions for this equation.
Answer: We don´t have any triangle with these measures.
Answer:
No triangle possible.
Step-by-step explanation:
Let in triangle ABC,
BC = a = 1 unit,
AC = b = 2 unit,
m∠A = 31°,
By the law of sine,
[tex]\frac{sin B}{b}=\frac{sin A}{a}[/tex]
[tex]\implies sin B=\frac{2sin 31^{\circ}}{1}=2sin 31^{\circ}=1.0301[/tex]
⇒ m∠B= undefined
( ∵ the value of sin Ф lies from 0 to 1 where Ф is an angle )
An angle of a triangle can not be undefined.
∴ Triangle ABC is not possible,
That is, there is no possible triangle with the given measures.
