Answer:
a) zero triangles.
b) one triangle.
Step-by-step explanation:
In triangle ABC:
- A, B and C are the interior angles.
- a, b and c are the sides opposite the corresponding interior angles.
Part (a)
Given:
- ∠A = 51°
- a = 10 cm
- b = 28 cm
Law of Sines
[tex]\sf \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)
To determine if any triangles are possible, substitute the given values into the Law of Sines to find angle B:
[tex]\implies \sf \dfrac{\sin 51^{\circ}}{10}=\dfrac{\sin B}{28}[/tex]
[tex]\implies \sf \sin B=\dfrac{28\sin 51^{\circ}}{10}[/tex]
[tex]\implies \sf \sin B=2.176008...[/tex]
As -1 ≤ sin B ≤ 1, there is no solution for angle B.
Therefore, zero triangles are possible.
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Part (b)
Given:
- ∠C = 30°
- a = 24 cm
- c = 12 cm
Law of Sines
[tex]\sf \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)
To determine if any triangles are possible, substitute the given values into the Law of Sines to find angle A:
[tex]\implies \sf \dfrac{\sin A}{24}=\dfrac{\sin 30^{\circ}}{12}[/tex]
[tex]\implies \sf \sin A=\dfrac{24\sin 30^{\circ}}{12}[/tex]
[tex]\implies \sf \sin A=1[/tex]
[tex]\implies \sf A=\sin^{-1}(1)[/tex]
[tex]\implies \sf A=90^{\circ}[/tex]
Therefore, one triangle is possible (see attachment).
