Question

Please solve the law of sines ambiguous case question. Please explain the questiom concisely and succinctly since I get confused easily with too complicated steps.​

Answer 1

Answer:

Two triangles are produced.

C₁ ≈ 46.31° (2 d.p.)

A₁ ≈ 93.69° (2 d.p.)

a₁ ≈ 12.42 (2 d.p.)

C₂ ≈ 133.69° (2 d.p.)

A₂ ≈ 6.31° (2 d.p.)

a₂ ≈ 1.37 (2 d.p.)

Step-by-step explanation:

Given:

• b = 8
• c = 9
• B = 40°

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.

Sine Rule

$\sf \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$

(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:

$\implies \sf \dfrac{\sin 40^{\circ}}{8}=\dfrac{\sin C}{9}$

$\implies \sf \sin C= \dfrac{9\sin 40^{\circ}}{8}$

$\implies \sf C= \sin^{-1}\left(\dfrac{9\sin 40^{\circ}}{8}\right)$

$\implies \sf C=46.31^{\circ}\:\: (2\:d.p.)$

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:

$\implies \sf 180^{\circ}-46.31^{\circ}=133.69^{\circ}\:(2\:d.p.)$

Therefore:

• C₁ ≈ 46.31° (2 d.p.)
• C₂ ≈ 133.69° (2 d.p.)

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.

$\implies \sf A_1+B_1+C_1=180^{\circ}$

$\implies \sf A_1+40^{\circ}+46.31^{\circ}=180^{\circ}$

$\implies \sf A_1=93.69^{\circ}$

$\implies \sf A_2+B_2+C_2=180^{\circ}$

$\implies \sf A_2+40^{\circ}+133.69^{\circ}=180^{\circ}$

$\implies \sf A_2=6.31^{\circ}$

Finally, use the Sine Rule again to find the measure of side a for each triangle.

$\sf \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

$\implies \sf \dfrac{a_1}{\sin A_1}=\dfrac{b}{\sin B}$

$\implies \sf \dfrac{a_1}{\sin 93.69^{\circ}}=\dfrac{8}{\sin 40^{\circ}}$

$\implies \sf a_1=\dfrac{8\sin 93.69^{\circ}}{\sin 40^{\circ}}$

$\implies \sf a_1=12.42\:\:(2 \:d.p.)$

$\implies \sf \dfrac{a_2}{\sin A_2}=\dfrac{b}{\sin B}$

$\implies \sf \dfrac{a_2}{\sin 6.31^{\circ}}=\dfrac{8}{\sin 40^{\circ}}$

$\implies \sf a_2=\dfrac{8 \sin 6.31^{\circ}}{\sin 40^{\circ}}$

$\implies \sf a_2=1.37\:\:(2\:d.p.)$