Question

Find b, given that a = 20, angle A = 30°, and angle B = 45° in triangle ABC

Answer 1

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 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