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:


`(a)/(sinA) = (b)/(sinB)`

`(20)/(sin30^0) = (b)/(sin45^0)`

`(20)/( (1)/(2) ) = (b)/( ( √(2) )/(2) )`

`20* ( √(2) )/(2) = b* (1)/(2)`

`(20 √(2) )/(2) = (b)/(2)`

`2*b =2*20 √(2)`

`2b = 40 √(2)`

`b = (40 √(2) )/(2)`

`\boxed{b = 20 √(2) }`

Answer 2

b = 20√2

Explanation:

Given: ΔABC

a = 20 , m∠A = 30° and m∠B = 45°

To find: value of b.

We use Sine result, which state that

`(a)/(sin\,A)=(b)/(sin\,B)`

Substituting given values we, get

`(20)/(sin\,30^(\circ))=(b)/(sin\,45^(\circ))`

we know that
`sin\,30^(\circ)=(1)/(2)\:and\:sin\,45^(\circ)=(1)/(√(2))`, we get

`(20)/((1)/(2))=(b)/((1)/(√(2)))`

`20{*2}=b*√(2)`

`b*√(2)=40`

`b=(40)/(√(2))`

`b=20√(2)`

Therefore, b = 20√2