Consider the non-triangle below. Suppose that ∠ACB=103∘ and m∠BAC=44∘, and that y=50.5cm. What of the value of X?

We know that the sum of angles in a triangle equals 180 degrees. Therefore, ∠BAC+∠cba+∠ACB=180∘ ∠CBA=18∘−∠BAC−∠ACB ∠CBA=180∘−44∘−103∘ ∠CBA=33∘ By using law of sines in the given triangle, (sin∠BACx)=(sin∠CBAy) (sin⁡44∘x)=(sin⁡33∘50.5) x=sin⁡44∘sin⁡33∘⋅50.5 x=0.69460.5446⋅50.5 x=64.41

Resolved: "Consider the non-triangle below. Suppose that ∠ACB=103∘ and..."

High School
Geometry
Trigonometry
Non-right triangles and trigonometry

Bevan Mcdonald

Answered question

2021-01-16

Consider the non-triangle below. Suppose that $∠ A C B = 103^{\circ}$ and $m ∠ B A C = 44^{\circ}$ , and that $y = 50.5 c m$ . What of the value of X?

Answer & Explanation

brawnyN

Skilled2021-01-17Added 91 answers

We know that the sum of angles in a triangle equals 180 degrees. Therefore,

∠ B A C + ∠ c b a + ∠ A C B = 180^{\circ}

∠ C B A = 18^{\circ} - ∠ B A C - ∠ A C B

∠ C B A = 180^{\circ} - 44^{\circ} - 103^{\circ}

∠ C B A = 33^{\circ}

By using law of sines in the given triangle,

(\frac{\sin ∠ B A C}{x}) = (\frac{\sin ∠ C B A}{y})

(\frac{{\sin 44}^{\circ}}{x}) = (\frac{{\sin 33}^{\circ}}{50.5})

x = \frac{{\sin 44}^{\circ}}{{\sin 33}^{\circ}} \cdot 50.5

x = \frac{0.6946}{0.5446} \cdot 50.5

x = 64.41

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