Question

Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles.

Answer 1

Hello!

The answer is:

The first triangle is:

$A=59.6\°\\C=71.4\°\\c=17.6units$

The second triangle is:

$A=120.4\°\\C=10.6\°\\c=3.41units$

Why?

To solve the triangles, we must remember the Law of Sines form.

Law of Sines can be expressed by the following relationship:

$\frac{a}{Sin(A)}=\frac{b}{Sin(B)}=\frac{c}{Sin(C)}$

Where,

a, b, and c are sides of the triangle

A, B, and C are angles of the triangle.

We are given,

$B=49\°\\a=16\\b=14$

So, solving the triangles, we have:

- First Triangle:

Finding A, we have:

$\frac{a}{Sin(A)}=\frac{b}{Sin(B)}\\\\Sin(A)=a*\frac{Sin(B)}{b}=16*\frac{Sin(49\°)}{14}\\\\Sin^{-1}(Sin(A)=Sin^{-1}(16*\frac{Sin(49\°)}{14})\\\\A=59.6\°$

Finding C, we have:

Now, if the sum of all the interior angles of a triangle is equal to 180°, we have:

$A+B+C=180\°\\\\C=180-A-B\\\\C=180\°-59.6\°-49\°=71.4\°$

Finding c, we have:

Then, now that we know C, we need to look for "c":

$\frac{14}{Sin(49\°)}=\frac{c}{Sin(71.4\°)}\\\\c=\frac{14}{Sin(49\°)}*Sin(71.4\°)=17.58=17.6units$

So, the first triangle is:

$A=59.6\°\\C=71.4\°\\c=17.6units$

- Second Triangle:

Finding A, we have:

$\frac{a}{Sin(A)}=\frac{b}{Sin(B)}\\\\Sin(A)=a*\frac{Sin(B)}{b}=16*\frac{Sin(49\°)}{14}\\\\Sin^{-1}(Sin(A)=Sin^{-1}(16*\frac{Sin(49\°)}{14})\\\\A=59.6\°$

Now, since that there are two triangles that can be formed, (angle and its suplementary angle) there are two possible values for A, and we have:

$A=180\°-59.6\°=120.4\°$

Finding C, we have:

Then, if the sum of all the interior angles of a triangle is equal to 180°, we have:

$A+B+C=180\°\\\\C=180\°-A-B\\\\C=180\°-120.4\°-49\°=10.6\°$

Then, now that we know C, we need to look for "c".

Finding c, we have:

$\frac{14}{Sin(49\°)}=\frac{c}{Sin(10.6)}\\\\c=\frac{14}{Sin(49)\°}*Sin(10.6\°)=3.41units$

so, The second triangle is:

$A=120.4\°\\C=10.6\°\\c=3.41units$

Have a nice day!