Question

Given: △ABC, AB=5 square root 2

m∠A=45°, m∠C=30°

Find: BC and AC

Answer 1

Part a)
`BC=10\ units`

Part b)
`AC=13.66\ units`

Explanation:

step 1

Find the length side BC

Applying the law of sines

we know that

`(AB)/(sin(C))=(BC)/(sin(A))`

we have

`AB=5√(2)\ units`

`A=45^o`

`C=30^o`

substitute

`(5√(2))/(sin(30^o))=(BC)/(sin(45^o))`

solve for BC

`BC=(5√(2))/(sin(30^o))(sin(45^o))`

`BC=10\ units`

step 2

Find the measure of angle B

we know that

The sum of the interior angles in a triangle must be equal to 180 degrees

so

`m\angle A+m\angle B+m\angle C=180^o`

substitute the given values

`45^o+m\angle B+30^o=180^o`

`75^o+m\angle B=180^o`

`m\angle B=180^o-75^o`

`m\angle B=105^o`

step 3

Find the length side AC

Applying the law of sines

we know that

`(AB)/(sin(C))=(AC)/(sin(B))`

we have

`AB=5√(2)\ units`

`A=45^o`

`B=105^o`

substitute

`(5√(2))/(sin(30^o))=(AC)/(sin(105^o))`

solve for AC

`AC=(5√(2))/(sin(30^o))(sin(105^o))`

`AC=13.66\ units`