What is the approximate length of side AB in triangle ABC?
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Answer:
A) 14 units
Step-by-step explanation:
we know that the measure of angle A = 80
Law of Sines:
24/sin 80 = AB/sin 35
cross-multiply to get:
sin 80·AB = sin 35·24
sin 80·AB = 13.765
AB = 13.765/sin 80
AB is approximately 14
The approximate length of side AB in triangle ABC is 14 units. The correct option is A.
The sine law of trigonometry helps us to equate the side of the triangles to the angles of the triangles. It is given by the formula,
[tex]\dfrac{Sin\ A}{\alpha} =\dfrac{Sin\ B}{\beta} =\dfrac{Sin\ C}{\gamma}[/tex]
where Sin A is the angle and α is the length of the side of the triangle opposite to angle A,
Sin B is the angle and β is the length of the side of the triangle opposite to angle B,
Sin C is the angle and γ is the length of the side of the triangle opposite to angle C.
In the given triangle, the measure of ∠A can be written as,
∠A + ∠B + ∠C = 180°
∠A + 65° + 35° = 180°
∠A = 80°
Now, The approximate length of side AB in triangle ABC can be found by using the sine law can be written as,
[tex]\dfrac{\sin A}{BC} = \dfrac{\sin B}{AC} = \dfrac{\sin C}{AB}[/tex]
The ratio now for side AB can be written as,
[tex]\dfrac{\sin A}{BC} = \dfrac{\sin C}{AB}[/tex]
AB = (sin C × AB)/(sin A)
AB = (sin35° × 24)/(sin 80°)
AB = 13.978 ≈ 14 units
Hence, the approximate length of side AB in triangle ABC is 14 units.
Learn more about Sine Law here:
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