Respuesta :
Recall the following formula:
Given a triangle with side lengths a, b, and c. Let the measure of the angle between sides of length a and c be B.
Then, the area of the triangle is given by :
[tex]Area= \frac{1}{2}\cdot a\cdot c \cdot \sin B[/tex].
In our example we have: Area=10 square cm, a=8 cm, c=5 cm, and we want to find the measure of the angle between the 2 sides a and c.
Substituting in the formula we have:
[tex]10= \frac{1}{2}\cdot 8\cdot 5 \cdot \sin B [/tex]
Thus, [tex]10=20 \sin B[/tex], which means [tex]\sin B = \frac{1}{2} [/tex].
[tex] \frac{1}{2} [/tex] is the sine of [tex]30^{\circ}[/tex], but also [tex]150^{\circ}[/tex].
Answer: both 30, and 150 degrees are possible
Given a triangle with side lengths a, b, and c. Let the measure of the angle between sides of length a and c be B.
Then, the area of the triangle is given by :
[tex]Area= \frac{1}{2}\cdot a\cdot c \cdot \sin B[/tex].
In our example we have: Area=10 square cm, a=8 cm, c=5 cm, and we want to find the measure of the angle between the 2 sides a and c.
Substituting in the formula we have:
[tex]10= \frac{1}{2}\cdot 8\cdot 5 \cdot \sin B [/tex]
Thus, [tex]10=20 \sin B[/tex], which means [tex]\sin B = \frac{1}{2} [/tex].
[tex] \frac{1}{2} [/tex] is the sine of [tex]30^{\circ}[/tex], but also [tex]150^{\circ}[/tex].
Answer: both 30, and 150 degrees are possible
