Question

How do you solve a triangle ABC where b=125 c=162 B=40 degrees?

Answer 1

Angle A = 83.59° (2 d.p.)

Angle C = 56.41° (2 d.p.)

Side a = 193.25 (2 d.p.)

Explanation:

`\boxed{\begin{minipage}{7.6 cm}\underline{Sine Rule} \\\\$(\sin A)/(a)=(\sin B)/(b)=(\sin C)/(c) $\\\\\\where:\\ \phantom{ww}$\bullet$ $A, B$ and $C$ are the angles. \\ \phantom{ww}$\bullet$ $a, b$ and $c$ are the sides opposite the angles.\\\end{minipage}}`

Given:

• b = 125
• c = 162
• B = 40°

Substitute the given values into the Sine Rule formula:

`\implies (\sin A)/(a)=(\sin 40^(\circ))/(125)=(\sin C)/(162)`

Solve for angle C:

`\implies (\sin 40^(\circ))/(125)=(\sin C)/(162)`

`\implies \sin C=(162\sin 40^(\circ))/(125)`

`\implies C=\sin^(-1)\left((162\sin 40^(\circ))/(125)\right)`

`\implies C=56.4136175...^(\circ)`

Interior angles of a triangle sum to 180°. Therefore:

`\implies A+B+C = 180^(\circ)`

`\implies A = 180^(\circ)-B-C`

`\implies A = 180^(\circ)-40^(\circ)-56.4136175...^(\circ)`

`\implies A = 83.5863824...^(\circ)`

Finally, to find a, substitute the found angles and sides into the Sine Rule and solve for a:

`\implies (\sin A)/(a)=(\sin B)/(b)`

`\implies (\sin 83.5863824...^(\circ))/(a)=(\sin 40^(\circ))/(125)`

`\implies a=(125\sin 83.5863824...^(\circ))/(\sin 40^(\circ))`

`\implies a=193.248396...`