Answer:
62
Step-by-step explanation:
Khan Academy. I got this correct, so trust me :)
The measurement of the angle A for the considered triangle is found being of 62° approximately.
For any triangle ABC, with side measures |BC| = a. |AC| = b. |AB| = c,
we have, by law of sines,
[tex]\dfrac{sin\angle A}{a} = \dfrac{sin\angle B}{b} = \dfrac{sin\angle C}{c}[/tex]
Remember that we took
[tex]\dfrac{\sin(angle)}{\text{length of the side opposite to that angle}}[/tex]
For the considered triangle, we can use sine rule.
In triangle ABC, we're given that:
The side opposite to angle A is BC and the side opposite to angle B is AC. Thus, by sine law, we get:
[tex]\dfrac{\sin(m\angle A)}{|BC|} = \dfrac{sin(m\angle B)}{|AC|}\\\\\dfrac{\sin(m\angle A)}{15} = \dfrac{\sin(28^\circ)}{8}\\\\\sin(m\angle A ) \approx \dfrac{15 \times 0.46947}{8} \approx 0.88026\\\\m\angle A \approx \sin^{-1}(0.88026)\\\\m\angle A \approx 61.67^\circ, 118.32^\circ[/tex]
But since angle A is acute(angle with measure of less than 90 degrees), thus, m∠A ≈ 61.67° ≈ 62°
Thus, the measurement of the angle A for the considered triangle is found being of 62° approximately.
Learn more about law of sines here:
https://brainly.com/question/17289163
