BC = 8.60, CD = 5.00, BD = 7.00
∠D = 90° , ∠C = 54° , ∠B = 36°
Explanation:
B(-2, 4), C(3, 3), D(-2, 3)
To get the three side lengths, we would find the distance between them.
Distance formula:
[tex]dis\tan ce\text{ = }\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2}[/tex][tex]\begin{gathered} x_1=-2,y_1=-4,x_2=3,y_2\text{ =3} \\ Distance\text{ BC = }\sqrt[]{(3-(-4))^2+(3-(-2))^2} \\ \text{Distance BC = }\sqrt[]{(3+4)^2+(3+2)^2}\text{ =}\sqrt[]{(7)^2+5^2\text{ }} \\ \text{Distance BC = }\sqrt[]{49+25}\text{ } \\ \text{Distance BC = }\sqrt[]{74}\text{ = 8}.60 \\ \text{Distance BC = 8.60 (nearest hundredth)} \end{gathered}[/tex][tex]\begin{gathered} x_1=3,y_1=3,x_2=-2,y_2\text{ =3} \\ \text{Distance CD = }\sqrt[]{(3-3)^2+(-2-3)^2} \\ \text{Distance CD =}\sqrt[]{(0)^2+(-5)^2}\text{ =}\sqrt[]{0+25} \\ \text{Distance CD =}\sqrt[]{25} \\ \text{ Distance CD = 5.00 (nearest hundredth)} \end{gathered}[/tex][tex]\begin{gathered} x_1=-2,y_1=-4,x_2=-2,y_2\text{ = 3} \\ \text{Distance BD = }\sqrt[]{(3-(-4))^2+(-2-\mleft(-2\mright))^2} \\ \text{Distance BD =}\sqrt[]{(3+4)^2+(-2+2)^2} \\ \text{Distance BD =}\sqrt[]{7^2+0^2}\text{ =}\sqrt[]{49} \\ \text{Distance BD = 7} \\ \text{Distance BD = 7.00 (nearest hundredth)} \end{gathered}[/tex]
From the diagram, we can see it is a right angled triangle.
∠D = 90°
To get angle C, we would apply the trigonometry: SOHCAHTOA
opposite = side opposite angle C = BD = 7
hypotenuse = BC = 8.60
sin C = opposite/hypotenuse
sin C = 7/8.6
sinC = 0.81395
C = arc sin (0.81395)
C = 54.48°
To the nearest degree = 54°
To get angle A:
angle B + angle D + angle C = 180° (sum of angles in a triangle)
angle B + 90 + 54 = 180
angle B + 144 = 180
angle B = 180 - 144
angle B = 36° (nearest degree)
