Question

Marty is asked to draw triangles with side lengths of 4 units and 2 units, and a non-included angle of 30°. Select all the triangles that he could have drawn. *

Answer 1

Answer:

The drawn in the attached figure

see the explanation

Step-by-step explanation:

First case

In the triangle ABC

Let

$a=4\ units\\b=2/ units\\B=30^o$

Applying the law of sines

Find the measure of angle A

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

substitute the given values

$\frac{4}{sin(A)}=\frac{2}{sin(30^o)}$

$sin(A)=1$

so

$A=90^o$

Find the measure of angle C

In a right triangle

we know that

$B+C=90^o$ ----> by complementary angles

$B=30^o$

therefore

$C=60^o$

Find the length side c

Applying the law of sines

$\frac{c}{sin(C)}=\frac{b}{sin(B)}$

substitute the given values

$\frac{c}{sin(60^o)}=\frac{2}{sin(30^o)}$

$c=2\sqrt{3}\ units$

therefore

The dimensions of the triangle are

$A=90^o$

$B=30^o$

$C=60^o$

$a=4\ units\\b=2\ units\\c=2\sqrt{3}=3.46\ units$

Second case

In the triangle ABC

Let

$a=4\ units\\b=2/ units\\A=30^o$

Applying the law of sines

Find the measure of angle B

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

substitute the given values

$\frac{4}{sin(30^o)}=\frac{2}{sin(B)}$

$sin(B)=0.25$

so

using a calculator

$B=14.48^o$

Find the measure of angle C

we know that

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

so

$A+B+C=180^o$

$A=30^o\\B=14.48^o$

therefore

$30^o+14.48^o+C=180^o$

$C=135.52^o$

Find the length side c

Applying the law of sines

$\frac{c}{sin(C)}=\frac{a}{sin(A)}$

substitute the given values

$\frac{c}{sin(135.52^o)}=\frac{4}{sin(30^o)}$

$c=5.61\ units$

therefore

The dimensions of the triangle are

$A=30^o$

$B=14.48^o$

$C=135.52^o$

$a=4\ units\\b=2\ units\\c=5.61\ units$

see the attached figure to better understand the problem