Respuesta :

msm555

Answer:

solution given;

let

AB=a

AC=b=30ft

AB=c=20ft

<A=115°

By using Cosine rule.

a²=b²+c²-2bc cos angle

a²=30²+20²-2*30*20 Cos 115°

a²=1807.1419

a=√[1807.1419]

a=42.51

Side A is 42.51ft.

Again

Cos B=[tex] \frac{a²+c²-b²}{2ac} [/tex]

Cos B=[tex] \frac{42.51²+20²-30²}{2*42.51*20} [/tex]

Cos B=0.7687

<B=Cos -¹(0.7687)

<B=39.46°

Angle B is 39.46

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To find [tex] \tt{\overline{A}}[/tex]

Given that:

[tex]\quad\quad\quad\quad\tt{ \angle{A = 115° }}[/tex]

Using cosine rule:

[tex] \tt{\overline{A}²=b²+c²-2bc \cos( \angle{a}) }[/tex]

[tex] \tt{\overline{A}²=(30)²+(20)²-2(30)(20) \cos( \angle{115°}) }[/tex]

[tex] \tt{\overline{A}²=900+400-2(600) \cos( \angle{115°}) }[/tex]

[tex] \tt{\overline{A}²=1300-1200 \cos( \angle{115°}) }[/tex]

[tex] \tt{\overline{A} ²=1807.1419}[/tex]

[tex] \tt{\overline{A}= \sqrt{ 1807.1419}}[/tex]

[tex] \pink {\boxed{ \tt{ \overline{A}=42.51}}}[/tex]

Now, to find [tex]\tt{ \angle{B}}[/tex]

[tex] \tt{ cos\;B = \frac{ {a}^{2} + {c}^{2} - {b}^{2} }{2ac} }[/tex]

[tex] \tt{ cos\:B = \frac{ {42.51}^{2} + {20}^{2} - {30}^{2} }{2(42.51)(20)} }[/tex]

[tex] \tt{ cos\:B = 0.7687}[/tex]

[tex] \tt{ \angle{B} = { \cos}^{ - 1} (0.7687)}[/tex]

[tex]\pink{\boxed{\tt{ \angle{B} = {39.46°}}}}[/tex]

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