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dexteright02

The side b is opposite to the angle B, applying the law of the sines, we have:

\(\frac{a}{sinA} = \frac{b}{sinB}\)
\(\frac{20}{sin30^0} = \frac{b}{sin45^0}\)
\(\frac{20}{ \frac{1}{2} } = \frac{b}{ \frac{ \sqrt{2} }{2} }\)
\(20* \frac{ \sqrt{2} }{2} = b* \frac{1}{2}\)
\(\frac{20 \sqrt{2} }{2} = \frac{b}{2}\)
\(2*b =2*20 \sqrt{2}\)
\(2b = 40 \sqrt{2}\)
\(b = \frac{40 \sqrt{2} }{2}\)
\(\boxed{b = 20 \sqrt{2} }\)

Answer:

b = 20√2

Step-by-step explanation:

Given: ΔABC

           a = 20 , m∠A =  30° and m∠B = 45°

To find: value of b.

We use Sine result, which state that

\(\frac{a}{sin\,A}=\frac{b}{sin\,B}\)

Substituting given values we, get

\(\frac{20}{sin\,30^{\circ}}=\frac{b}{sin\,45^{\circ}}\)

we know that \(sin\,30^{\circ}=\frac{1}{2}\:and\:sin\,45^{\circ}=\frac{1}{\sqrt{2}}\), we get

\(\frac{20}{\frac{1}{2}}=\frac{b}{\frac{1}{\sqrt{2}}}\)

\(20{\times2}=b\times\sqrt{2}\)

\(b\times\sqrt{2}=40\)

\(b=\frac{40}{\sqrt{2}}\)

\(b=20\sqrt{2}\)

Therefore, b = 20√2

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