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Solve ΔABC using the Law of Cosines

1. B= 36°, c = 19, a = 11

2. a = 21, b = 26, c = 17

Answer:

1)  A = 32.6°, C = 111.4°, b = 12.0

2) A = 53.6°, B = 85.7°, C = 40.7°

Step-by-step explanation:

Given values of triangle ABC:

First, find the measure of side b using the Law of Cosines for finding sides.

[tex]\boxed{\begin{minipage}{6 cm}\underline{Law of Cosines (for finding sides)} \\\\$c^2=a^2+b^2-2ab \cos (C)$\\\\where:\\ \phantom{ww}$\bullet$ $a, b$ and $c$ are the sides.\\ \phantom{ww}$\bullet$ $C$ is the angle opposite side $c$. \\\end{minipage}}[/tex]

As the given angle is B, change C for B in the formula and swap b and c:

[tex]b^2=a^2+c^2-2ac\cos(B)[/tex]

Substitute the given values and solve for b:

[tex]\implies b^2=11^2+19^2-2(11)(19)\cos(36^{\circ})[/tex]

[tex]\implies b^2=482-418\cos(36^{\circ})[/tex]

[tex]\implies b=\sqrt{482-418\cos(36^{\circ})}[/tex]

[tex]\implies b=11.9929519...[/tex]

Now we have the measures of all three sides of the triangle, we can use the Law of Cosines for finding angles to find the measures of angles A and C.

[tex]\boxed{\begin{minipage}{7.6 cm}\underline{Law of Cosines (for finding angles)} \\\\$\cos(C)=\dfrac{a^2+b^2-c^2}{2ab}$\\\\\\where:\\ \phantom{ww}$\bullet$ $C$ is the angle. \\ \phantom{ww}$\bullet$ $a$ and $b$ are the sides adjacent the angle. \\ \phantom{ww}$\bullet$ $c$ is the side opposite the angle.\\\end{minipage}}[/tex]

To find the measure of angle A, swap a and c in the formula, and change C for A:

[tex]\implies \cos(A)=\dfrac{c^2+b^2-a^2}{2cb}[/tex]

[tex]\implies \cos(A)=\dfrac{19^2+(11.9929519...)^2-11^2}{2(19)(11.9929519...)}[/tex]

[tex]\implies \cos(A)=0.842229094...[/tex]

[tex]\implies A=\cos^{-1}(0.842229094...)[/tex]

[tex]\implies A=32.6237394...^{\circ}[/tex]

To find the measure of angle C, substitute the values of a, b and c into the formula:

[tex]\implies \cos(C)=\dfrac{a^2+b^2-c^2}{2ab}[/tex]

[tex]\implies \cos(C)=\dfrac{11^2+(11.9929519...)^2-19^2}{2(11)(11.9929519...)}[/tex]

[tex]\implies \cos(C)=-0.364490987...[/tex]

[tex]\implies C=\cos^{-1}(-0.364490987...)[/tex]

[tex]\implies C=111.376260...^{\circ}[/tex]

Therefore, the remaining side and angles for triangle ABC are:

[tex]\hrulefill[/tex]

To solve for the remaining angles of the triangle ABC given its side lengths, use the Law of Cosines for finding angles.

[tex]\boxed{\begin{minipage}{7.6 cm}\underline{Law of Cosines (for finding angles)} \\\\$\cos(C)=\dfrac{a^2+b^2-c^2}{2ab}$\\\\\\where:\\ \phantom{ww}$\bullet$ $C$ is the angle. \\ \phantom{ww}$\bullet$ $a$ and $b$ are the sides adjacent the angle. \\ \phantom{ww}$\bullet$ $c$ is the side opposite the angle.\\\end{minipage}}[/tex]

Given sides of triangle ABC:

Substitute the values of a, b and c into the Law of Cosines formula and solve for angle C:

[tex]\implies \cos(C)=\dfrac{a^2+b^2-c^2}{2ab}[/tex]

[tex]\implies \cos(C)=\dfrac{21^2+26^2-17^2}{2(21)(26)}[/tex]

[tex]\implies \cos(C)=\dfrac{828}{1092}[/tex]

[tex]\implies C=\cos^{-1}\left(\dfrac{828}{1092}\right)[/tex]

[tex]\implies C=40.690560...^{\circ}[/tex]

To find the measure of angle B, swap b and c in the formula, and change C for B:

[tex]\implies \cos(B)=\dfrac{a^2+c^2-b^2}{2ac}[/tex]

[tex]\implies \cos(B)=\dfrac{21^2+17^2-26^2}{2(21)(17)}[/tex]

[tex]\implies \cos(B)=\dfrac{54}{714}[/tex]

[tex]\implies B=\cos^{-1}\left(\dfrac{54}{714}\right)[/tex]

[tex]\implies B=85.6625640...^{\circ}[/tex]

To find the measure of angle A, swap a and c in the formula, and change C for A:

[tex]\implies \cos(A)=\dfrac{c^2+b^2-a^2}{2cb}[/tex]

[tex]\implies \cos(A)=\dfrac{17^2+26^2-21^2}{2(17)(26)}[/tex]

[tex]\implies \cos(A)=\dfrac{524}{884}[/tex]

[tex]\implies A=\cos^{-1}\left(\dfrac{524}{884}\right)[/tex]

[tex]\implies A=53.6468753...^{\circ}[/tex]

Therefore, the measures of the angles of triangle ABC with sides a = 21, b = 26 and c = 17 are:

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