Notes

Use the distance formula to find the side lengths. Enter each length to the nearest tenth.
AB = (3 − (−2))2 + (−3 + 2)2
= 26
BC = (3 − (3))2 + (3 + 3)2
= 36
= 6
AC = (3 − (−2))2 + (3 + 2)2
= 50
= 52
Enter the perimeter of the triangle.
Perimeter = BC + AB + AC
= 6 + 26 + 52
Enter the area of the triangle. Let the base b of the triangle be equal to BC = 6. The height is the shortest distance between the base BC and point A, which is h = 5.
Area =
1
2
bh
Area =
1
2
(6)(5) Substitute 6 for b and 5 for h.
Area = 15 Evaluate.
Use the value of the area to find ∠C.
Area =
1
2
absinC
Substitute the two adjacent sides of ∠C for a and b.
15 =
1
2
(6)(52)sinC
15 = 152sinC
= sinC
45° = m∠C
Use the area formula to find ∠A.
Area =
1
2
bcsinA
Substitute the two adjacent sides of ∠A for b and c.
15 =
1
2
(26)(52)sinA
15 = 513sinA
= sinA
= m∠A
56° ≈ m∠A
The sum of all angles in a triangle is equal to 180°.
m∠A + m∠B + m∠C = 180°
56° + m∠B + 45° = 180° Substitute 56 for m∠A and 45 for m∠C.
m∠B ≈ 79° Subtract 56 and 45 from both sides and solve.

Question 18 of 19