Question

List the angles of the triangle in order from smallest to largest. In triangle Upper A Upper C Upper B, line segment Upper A Upper C has length 4.3, line segment Upper B Upper C has length 5.9, and line segment Upper B Upper A has length 3.5. A C B 5.9 4.3 3.5 Choose the correct order of the angles from smallest to largest.

Answer 1

Answer:

$C = 36.0$

$B = 46.2$

$A = 97.8$

Step-by-step explanation:

Given

$\triangle ABC$

$AC = 4.3$

$BC = 5.9$

$BA = 3.5$

Required

List the angles from smallest to largest

The given parameters is illustrated with the attached image.

$AC = 4.3$ -- b

$BC = 5.9$ --- a

$BA = 3.5$ --- c

This question will be solved using cosine rule

To calculate A, we have:

$a^2 = b^2 + c^2 -2bc\ cos(A)$

So, we have:

$5.9^2 = 4.3^2 + 3.5^2 - 2 * 4.3 * 3.5 * \cos(A)$

$34.81 = 18.49+ 12.25 - 30.10* \cos(A)$

Collect like terms

$34.81 - 18.49- 12.25 = - 30.10* \cos(A)$

$4.07 = - 30.10* \cos(A)$

Make cos(A) the subject

$\cos(A) = -\frac{4.07}{30.10}$

$\cos(A) = -0.1352$

Take arccos of both sides

$A = cos^{-1}(-0.1352)$

$A = 97.8$

Solving for B, we have:

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

This gives:

$4.3^2 = 5.9^2 + 3.5^2 -2*5.9*3.5\ cos(B)$

$18.49 = 34.81+ 12.25 -41.30 *\cos(B)$

Collect like terms

$18.49 - 34.81 - 12.25 = -41.30 *\cos(B)$

$-28.57 = -41.30 *\cos(B)$

Solve for cos(B)

$\cos(B) = \frac{-28.57}{-41.30}$

$\cos(B) = 0.6918$

Take arccos of both sides

$B = cos^{-1}(0.6918)$

$B = 46.2$

To solve for C, we make use of:

$A + B + C = 180$ --- angles in a triangle

$97.8 + 46.2 + C = 180$

Collect like terms

$C = - 97.8 - 46.2 + 180$

$C = 36.0$

So, we have:

$C = 36.0$

$B = 46.2$

$A = 97.8$