+5265 Right Triangle ABC is graphed on the coordinate plane and has vertices A(-1, 3), B(0, 5), and C(4, 3). What is the measure of angle C to the nearest degree?
F 27°
G 29°
H 32°
J 43°
Please hurry! Thanks!
+128570 Let's get the lengths of the sides of the triangle, first. Then, we can use the Law of Sines to find C.
A(-1, 3), B(0, 5), and C(4, 3)
BC =.sqrt [ (4 - 0)^2 + (5-3)^2 ] = sqrtt [16 + 4] = sqrt(20)
AC = sqrt [ (-1- 4)^2 + ( 3 - 3)^2]= sqrt [ 25] = 5
AB =sqrt [ (-1 - 0)^2 + ( 5 -3)^2 ] = sqrt (5)
So....AC is the hypotenuse.......and we're looking for C....so....using the Law of Sines
sin (90) / AC = sin C / sqrt(5)
1/5 = sinC / sqrt(5)
sinC = sqrt(5) / 5 = 1 / sqrt(5)
And using the sine inverse
arcsin ( 1 / sqrt(5) = C = about 26.57° = 27° [rounded]
+128570 Let's get the lengths of the sides of the triangle, first. Then, we can use the Law of Sines to find C.
A(-1, 3), B(0, 5), and C(4, 3)
BC =.sqrt [ (4 - 0)^2 + (5-3)^2 ] = sqrtt [16 + 4] = sqrt(20)
AC = sqrt [ (-1- 4)^2 + ( 3 - 3)^2]= sqrt [ 25] = 5
AB =sqrt [ (-1 - 0)^2 + ( 5 -3)^2 ] = sqrt (5)
So....AC is the hypotenuse.......and we're looking for C....so....using the Law of Sines
sin (90) / AC = sin C / sqrt(5)
1/5 = sinC / sqrt(5)
sinC = sqrt(5) / 5 = 1 / sqrt(5)
And using the sine inverse
arcsin ( 1 / sqrt(5) = C = about 26.57° = 27° [rounded]
