Question

NO LINKS!!! URGENT HELP PLEASE!!!!

Solve ΔABC using the Law of Cosines part 1

3. a = 12, b = 13, c = 20

4. A = 78°, b = 18, c = 10

Answer 1

3) A = 35.2°, B = 38.6°, C = 106.2°

4) B = 70.4°, C = 31.6°, a = 18.7

Explanation:

Question 3

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

`\boxed{\begin{minipage}{7.6 cm}\underline{Law of Cosines (for finding angles)} \\\\$\cos(C)=(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}}`

Given sides of triangle ABC:

• a = 12
• b = 13
• c = 20

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

`\implies \cos(C)=(a^2+b^2-c^2)/(2ab)`

`\implies \cos(C)=(12^2+13^2-20^2)/(2(12)(13))`

`\implies \cos(C)=(-87)/(312)`

`\implies \cos(C)=-(29)/(104)`

`\implies C=\cos^(-1)\left(-(29)/(104)\right)`

`\implies C=106.191351...^(\circ)`

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

`\implies \cos(B)=(a^2+c^2-b^2)/(2ac)`

`\implies \cos(B)=(12^2+20^2-13^2)/(2(12)(20))`

`\implies \cos(B)=(375)/(480)`

`\implies B=\cos^(-1)\left((375)/(480)\right)`

`\implies B=38.6248438...^(\circ)`

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

`\implies \cos(A)=(c^2+b^2-a^2)/(2cb)`

`\implies \cos(A)=(20^2+13^2-12^2)/(2(20)(13))`

`\implies \cos(A)=(425)/(520)`

`\implies A=\cos^(-1)\left((425)/(520)\right)`

`\implies A=35.1838154...^(\circ)`

Therefore, the measures of the angles of triangle ABC with sides a = 12, b = 13 and c = 20 are:

• A = 35.2°
• B = 38.6°
• C = 106.2°

`\hrulefill`

Question 4

Given values of triangle ABC:

• A = 78°
• b = 18
• c = 10

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

`\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}}`

As the given angle is A, change C for A in the formula and swap a and c:

`\implies a^2=c^2+b^2-2cb \cos (A)`

Substitute the given values and solve for a:

`\implies a^2=10^2+18^2-2(10)(18) \cos (78^(\circ))`

`\implies a^2=424-360\cos (78^(\circ))`

`\implies a=\sqrt{424-360\cos (78^(\circ))}`

`\implies a=18.6856038...`

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 B and C.

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

`\implies \cos(C)=(a^2+b^2-c^2)/(2ab)`

`\implies \cos(C)=((18.6856038...)^2+18^2-10^2)/(2(18.6856038...)(18))`

`\implies \cos(C)=0.852040063...`

`\implies C=31.565743...^(\circ)`

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

`\implies \cos(B)=(a^2+c^2-b^2)/(2ac)`

`\implies \cos(B)=((18.6856038...)^2+10^2-18^2)/(2(18.6856038...)(10))`

`\implies cos(B)=0.334888270...`

`\implies B=\cos^(-1)(0.334888270...)`

`\implies B=70.434256...^(\circ)`

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

• B = 70.4°
• C = 31.6°
• a = 18.7