Question

Solve the following triangles. Identify the case # in each given
triangle.

6.) Given parts:

a= 173 b= 267 c=412

Answer 1

Here is the sketch of the triangle;

We are given all the three sides. We need to calculate the three angles A, B, and C.

Here we apply cosine rule.

$a^2 = b^2+c^2 - 2bccosA$

And this can be re-written as;

$cosA = \frac{b^2 + c^2 - a^2}{2bc}$

$cos A = \frac{267^2+ 412^2 - 173^2}{2 × 267 × 412}$

$=\frac{71289+169744-29929}{220008}$

$=\frac{211104}{220008}$

$= 0.95952874441$

$\therefore cosA =0.95952874441$

$A = cos^{-1} (0.95952874441)$

$= 16.36 °$

To find angle B, we again apply cosine rule

$b^2= a^2 +c^2 - 2ac cos B$

Which can also be re-written;

$cos B= \frac{a^2+c^2-b^2}{2ac}$

$cos B= \frac{173^2+412^2-267^2}{2 × 173 × 412}$

$=\frac{29929+169744-71289}{142552}$

$=\frac{128384}{142552}$

$=0.90061170661$

$B = cos^{-1}(0.90061170661)$

$B= 25.76 °$

To find the final angle C, we apply the property: 'The sum of the three interior angles of a triangle add up to 180°'

$\therefore C = 180 ° - (A+B)$

$= 180 ° - (16.36 ° + 25.76 °)$

$=180 ° - 42.12 °$

$=137.88 °$

Therefore, the three angles are;

$A= 16.36 °$

$B = 25.76 °$

$C=137.88 °$