How do you determine whether $△ A B C$ $triangle ABC$ has no, one, or two solutions given $A = 75^{\circ}, a = 14, b = 11$ $A=75^circ, a=14, b=11$ ?

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How do you determine whether $△ A B C$ $triangle ABC$ has no, one, or two solutions given $A = 75^{\circ}, a = 14, b = 11$ $A=75^circ, a=14, b=11$ ?

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Answer 1 · 2017-10-21T18:46:27

$A = 75^{o}, B = {40.63}^{o}, C = {64.37}^{o}$ $A=75^o, B= 40.63^o, C = 64.37^o$

$a = 14, b = 11, c = 13.07$ $a =14, b=11, c= 13.07$

Explanation:

I'm not sure what you mean by no, one or two solutions. We can solve the triangle completely with the given information. So in that respect it will have one solution.

Using Sine Rule:

$\frac{sin (A)}{a} = \frac{sin (B)}{b} = \frac{sin (C)}{c}$ $sin(A)/a=sin(B)/b=sin(C)/c$

$\frac{sin (75)}{14} = \frac{sin (B)}{11} \Rightarrow sin (B) = \frac{11 sin (75)}{14}$ $sin(75)/14=sin(B)/11=>sin(B)= (11sin(75))/14$

$B = arcsin (\frac{11 sin (75)}{14}) = {40.63}^{o}$ $B= arcsin((11sin(75))/14)=40.63^o$ (2 .d.p.)

$C = 180 - (75 + 40.63) = {64.37}^{o}$ $C = 180-(75+40.63)=64.37^o$

$\frac{sin (75)}{14} = \frac{sin (64.37)}{c} \Rightarrow c = \frac{14 sin (64.37)}{sin (75)} = 13.07$ $sin(75)/14=sin(64.37)/c=>c=(14sin(64.37))/sin(75)=13.07$ (2 .d.p.)

So:

$A = 75^{o}, B = {40.63}^{o}, C = {64.37}^{o}$ $A=75^o, B= 40.63^o, C = 64.37^o$

$a = 14, b = 11, c = 13.07$ $a =14, b=11, c= 13.07$

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How do you determine whether $△ A B C$ $triangle ABC$ has no, one, or two solutions given $A = 75^{\circ}, a = 14, b = 11$ $A=75^circ, a=14, b=11$ ?

1 Answer

Explanation:

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Explanation:

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How do you determine whether $△ A B C$ $triangle ABC$ has no, one, or two solutions given $A = 75^{\circ}, a = 14, b = 11$ $A=75^circ, a=14, b=11$ ?