Question

1) Solve for Side A.

2) Solve for Angle B.

Answer 1

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To find
`\tt{\overline{A}}`

Given that:

`\quad\quad\quad\quad\tt{ \angle{A = 115° }}`

Using cosine rule:

`\tt{\overline{A}²=b²+c²-2bc \cos( \angle{a}) }`

`\tt{\overline{A}²=(30)²+(20)²-2(30)(20) \cos( \angle{115°}) }`

`\tt{\overline{A}²=900+400-2(600) \cos( \angle{115°}) }`

`\tt{\overline{A}²=1300-1200 \cos( \angle{115°}) }`

`\tt{\overline{A} ²=1807.1419}`

`\tt{\overline{A}= √( 1807.1419)}`

`\pink {\boxed{ \tt{ \overline{A}=42.51}}}`

Now, to find
`\tt{ \angle{B}}`

`\tt{ cos\;B = \frac{ {a}^(2) + {c}^(2) - {b}^(2) }{2ac} }`

`\tt{ cos\:B = \frac{ {42.51}^(2) + {20}^(2) - {30}^(2) }{2(42.51)(20)} }`

`\tt{ cos\:B = 0.7687}`

`\tt{ \angle{B} = { \cos}^( - 1) (0.7687)}`

`\pink{\boxed{\tt{ \angle{B} = {39.46°}}}}`

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Answer 2

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=
`(a²+c²-b²)/(2ac)`

Cos B=
`(42.51²+20²-30²)/(2*42.51*20)`

Cos B=0.7687

<B=Cos -¹(0.7687)

<B=39.46°

Angle B is 39.46