Question

Given: △ABC, AB=5 square root 2 m∠A=45°, m∠C=30°Find: BC and AC

Answer 1

Answer:

Part a) $BC=10\ units$

Part b) $AC=13.66\ units$

Step-by-step explanation:

step 1

Find the length side BC

Applying the law of sines

we know that

$\frac{AB}{sin(C)}=\frac{BC}{sin(A)}$

we have

$AB=5\sqrt{2}\ units$

$A=45^o$

$C=30^o$

substitute

$\frac{5\sqrt{2}}{sin(30^o)}=\frac{BC}{sin(45^o)}$

solve for BC

$BC=\frac{5\sqrt{2}}{sin(30^o)}(sin(45^o))$

$BC=10\ units$

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

$\frac{AB}{sin(C)}=\frac{AC}{sin(B)}$

we have

$AB=5\sqrt{2}\ units$

$A=45^o$

$B=105^o$

substitute

$\frac{5\sqrt{2}}{sin(30^o)}=\frac{AC}{sin(105^o)}$

solve for AC

$AC=\frac{5\sqrt{2}}{sin(30^o)}(sin(105^o))$

$AC=13.66\ units$