Question

Options: (50)^1/2, (65)^1/2, (105)^1/2, (145)^1/2

last sentence options: 55.21, 85.16, 105.26, 114.11

Answer 1

Explanation:

Vertices of ΔABC are,

A(-3, 6), B(2, 1) and C(9, 5)

Use the formula to get the distance between two points
`(x_1,y_1)` and
`(x_2,y_2)`,

Distance =
`√((x_2-x_1)^2+(y_2-y_1)^2)`

By using the formula,

AB =
`√((1-6)^2+(2+3)^2)`

=
`√(50)` units

BC =
`√((5-1)^2+(9-2)^2)`

=
`√(65)` units

AC =
`√((6-5)^2+(-3-9)^2)`

=
`√(145)`

Use cosine rule to find the measure of ∠ABC.

AC² = AB² + BC²- 2(AB)(BC)cos(B)

`(√(145))^2=(√(50))^2+(√(65))^2-2(√(50))(√(65))\text{cosB}`

145 = 50 + 65 - 2(√3250)cosB

cos(B) =
`-((145-115)/(2√(3250)))`

= -0.26312

B =
`\text{cos}^(-1)(-0.26312)`

B = 105.26°