Answer:
[tex] \large \bold{Two \: triangles \: are \: produced.} \\ [/tex]
[tex]\qquad \bold{C_1 \approx 46.31^{\circ} (2 d.p.)} \\ [/tex]
[tex]\qquad \bold{A_1 \approx 93.69^{\circ} (2 d.p.)} \\ [/tex]
[tex] \qquad\bold{a_1 \approx 12.42 (2 d.p.)} \\ [/tex]
[tex]\qquad \bold{C_2 \approx 133.69^{\circ}(2 d.p.)} \\ [/tex]
[tex]\qquad \bold{A_2 \approx 6.31^{\circ} (2 d.p.)} \\ [/tex]
[tex]\qquad \bold{a_2 \approx 1.37 (2 d.p.)} \\ [/tex]
Step-by-step explanation:
[tex] \\ \sf{Given:} \\ [/tex]
- [tex] \bold{b = 8} \\ [/tex]
- [tex] \bold{c = 9} \\ [/tex]
- [tex] \bold{B = 40^{\circ}} \\ [/tex]
Two triangles can be produced with the given information (see attached)
Angle C can be acute (less than 90°) and obtuse (greater than 90° and less than 180°).
Use the Sine Rule to find the measure of angle C.
[tex] \\ \large \underline{ \sf{ \blue{Sine \: Rule : }}} \\ [/tex]
[tex] \qquad \hookrightarrow \: \boxed{\bold{ \: \: \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c} \: \: } } \\ [/tex]
(where A, B and C are the angles and a, b and c are the sides opposite the angles).
Substitute the given information into the Sine Rule formula and solve for C:
[tex]\implies \: \sf \dfrac{\sin 40^{\circ}}{8}=\dfrac{\sin C}{9} \\ [/tex]
[tex]\implies \: \sf \sin C= \dfrac{9\sin 40^{\circ}}{8} \\ [/tex]
[tex]\implies \: \sf C= \sin^{-1}\left(\dfrac{9\sin 40^{\circ}}{8}\right) \\ [/tex]
[tex]\implies \: \sf C=46.31^{\circ}\:\: (2\:d.p.) \\ [/tex]
The found angle is acute and is therefore the measure of angle C₁.
If C₁ is acute, then its supplement is obtuse as sin θ = sin (180° - θ), where 90° < θ < 180°.
Therefore, angle C₂ is:
[tex]\implies \: \sf 180^{\circ}-46.31^{\circ}=133.69^{\circ}\:(2\:d.p.) \\ [/tex]
[tex] \large \sf{Therefore:} \\ [/tex]
- [tex] \bold{C_1 \approx 46.31^{\circ} (2 d.p.)} \\ [/tex]
- [tex] \bold{C_2 \approx 133.69^{\circ} (2 d.p.)} \\ [/tex]
Now we have found angles C₁ and C₂, use the theorem of interior angles of a triangle sum to 180° to find A₁ and A₂.
[tex]\implies \: \sf A_1+B_1+C_1=180^{\circ} \\ [/tex]
[tex]\implies \: \sf A_1+40^{\circ}+46.31^{\circ}=180^{\circ} \\ [/tex]
[tex]\implies \: \sf A_1=93.69^{\circ} \\ [/tex]
[tex]\implies \: \sf A_2+B_2+C_2=180^{\circ} \\ [/tex]
[tex]\implies \: \sf A_2+40^{\circ}+133.69^{\circ}=180^{\circ} \\ [/tex]
[tex]\implies \: \sf A_2=6.31^{\circ} \\ [/tex]
Finally, use the Sine Rule again to find the measure of side a for each triangle.
[tex]\sf \qquad \mapsto \: \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C} \\ [/tex]
[tex]\implies \: \sf \dfrac{a_1}{\sin A_1}=\dfrac{b}{\sin B} \\ [/tex]
[tex]\implies \: \sf \dfrac{a_1}{\sin 93.69^{\circ}}=\dfrac{8}{\sin 40^{\circ}} \\ [/tex]
[tex]\implies \: \sf a_1=\dfrac{8\sin 93.69^{\circ}}{\sin 40^{\circ}} \\ [/tex]
[tex]\implies \: \sf a_1=12.42\:\:(2 \:d.p.) \\ [/tex]
[tex]\implies \: \sf \dfrac{a_2}{\sin A_2}=\dfrac{b}{\sin B} \\ [/tex]
[tex]\implies \: \sf \dfrac{a_2}{\sin 6.31^{\circ}}=\dfrac{8}{\sin 40^{\circ}} \\ [/tex]
[tex]\implies \: \sf a_2=\dfrac{8 \sin 6.31^{\circ}}{\sin 40^{\circ}} \\ [/tex]
[tex]\implies \: \sf a_2=1.37\:\:(2\:d.p.) [/tex]
