Question

solve the triangle.
a = 11.0 ft, b = 13.0 ft, c = 22.0 ft
what is the measure of angle a?
a = □°
(round to two decimal places as needed.)
what is the measure of angle b?
b = □°
(round to two decimal places as needed.)
what is the measure of angle c?
c = □°
(round to two decimal places as needed.)

Explanation:

To solve for the angles of the triangle with sides \( a = 11.0 \) ft, \( b = 13.0 \) ft, and \( c = 22.0 \) ft, we use the Law of Cosines. The Law of Cosines states that for a triangle with sides \( a \), \( b \), and \( c \) opposite angles \( A \), \( B \), and \( C \) respectively, \( \cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc} \), \( \cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac} \), and \( \cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab} \). We can also use the fact that the sum of the angles in a triangle is \( 180^\circ \) to find the third angle after finding two.

Step 1: Find angle \( A \)

We use the Law of Cosines formula for angle \( A \):

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

Substitute \( a = 11.0 \), \( b = 13.0 \), and \( c = 22.0 \) into the formula:

$$ \cos A=\frac{(13.0)^{2}+(22.0)^{2}-(11.0)^{2}}{2\times13.0\times22.0} $$

First, calculate the numerator:

$$ (13.0)^{2}=169.0, \quad (22.0)^{2}=484.0, \quad (11.0)^{2}=121.0 $$

$$ \text{Numerator}=169.0 + 484.0-121.0=532.0 $$

Then, calculate the denominator:

$$ 2\times13.0\times22.0 = 572.0 $$

So,

$$ \cos A=\frac{532.0}{572.0}\approx0.9301 $$

Now, take the arccosine to find \( A \):

$$ A=\arccos(0.9301)\approx21.55^\circ $$

Step 2: Find angle \( B \)

We use the Law of Cosines formula for angle \( B \):

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

Substitute \( a = 11.0 \), \( b = 13.0 \), and \( c = 22.0 \) into the formula:

$$ \cos B=\frac{(11.0)^{2}+(22.0)^{2}-(13.0)^{2}}{2\times11.0\times22.0} $$

First, calculate the numerator:

$$ (11.0)^{2}=121.0, \quad (22.0)^{2}=484.0, \quad (13.0)^{2}=169.0 $$

$$ \text{Numerator}=121.0 + 484.0-169.0=436.0 $$

Then, calculate the denominator:

$$ 2\times11.0\times22.0 = 484.0 $$

So,

$$ \cos B=\frac{436.0}{484.0}\approx0.9008 $$

Now, take the arccosine to find \( B \):

$$ B=\arccos(0.9008)\approx25.74^\circ $$

Step 3: Find angle \( C \)

We know that the sum of the angles in a triangle is \( 180^\circ \). So,

$$ C = 180^\circ-(A + B) $$

Substitute \( A\approx21.55^\circ \) and \( B\approx25.74^\circ \):

$$ C=180^\circ-(21.55^\circ + 25.74^\circ)=180^\circ - 47.29^\circ = 132.71^\circ $$

We can also verify using the Law of Cosines:

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

Substitute \( a = 11.0 \), \( b = 13.0 \), and \( c = 22.0 \):

$$ \cos C=\frac{(11.0)^{2}+(13.0)^{2}-(22.0)^{2}}{2\times11.0\times13.0} $$

Calculate the numerator:

$$ 121.0+169.0 - 484.0=-194.0 $$

Denominator:

$$ 2\times11.0\times13.0 = 286.0 $$

$$ \cos C=\frac{-194.0}{286.0}\approx - 0.6783 $$

$$ C=\arccos(-0.6783)\approx132.71^\circ $$

Which matches the previous result.

Answer:

• Measure of angle \( A \): \( \boldsymbol{21.55^\circ} \)
• Measure of angle \( B \): \( \boldsymbol{25.74^\circ} \)
• Measure of angle \( C \): \( \boldsymbol{132.71^\circ} \)