Question

solve the triangle.
a = 11.0 ft, b = 12.0 ft, c = 22.0 ft
what is the measure of angle a?
a = \\(\square\\)°
(round to two decimal places as needed.)
what is the measure of angle b?
b = 17.73 °
(round to two decimal places as needed.)
what is the measure of angle c?
c = \\(\square\\)°
(round to two decimal places as needed.)

Explanation:

Step1: Check Triangle Inequality

First, check if the triangle is valid. The sum of any two sides must be greater than the third side. But \(11 + 12 = 23\), and \(23>22\); \(11 + 22 = 33>12\); \(12 + 22 = 34>11\). Wait, but actually, for a triangle, the sum of the two smaller sides should be greater than the largest side. Here, the two smaller sides are \(11\) and \(12\), and their sum is \(23\), which is greater than \(22\), so it is a valid triangle. Wait, but let's use the Law of Cosines.

Step2: Law of Cosines for Angle A

The Law of Cosines formula for angle \(A\) (opposite side \(a\)) is:

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

Substitute \(a = 11\), \(b = 12\), \(c = 22\):

$$ \cos A=\frac{12^{2}+22^{2}-11^{2}}{2\times12\times22}=\frac{144 + 484-121}{528}=\frac{507}{528}\approx0.9602 $$

Then, \(A=\arccos(0.9602)\approx15.66^{\circ}\)

Step3: Law of Cosines for Angle C

For angle \(C\) (opposite side \(c\)):

$$ \cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab} $$

Substitute \(a = 11\), \(b = 12\), \(c = 22\):

$$ \cos C=\frac{11^{2}+12^{2}-22^{2}}{2\times11\times12}=\frac{121 + 144-484}{264}=\frac{-219}{264}\approx - 0.8295 $$

Then, \(C=\arccos(-0.8295)\approx146.61^{\circ}\)

Step4: Check Angle Sum

The sum of angles in a triangle should be \(180^{\circ}\). Let's check: \(15.66+17.73 + 146.61=180^{\circ}\) (approximate, due to rounding).

Answer:

For angle \(A\): \(A\approx15.66^{\circ}\)
For angle \(C\): \(C\approx146.61^{\circ}\)