Question

How many unique triangles can be formed with two side lengths of 10 centimeters and one 40° angle

Answer 1

There are two unique triangles that can be formed two side lengths of 10 centimeters and one 40º angle.

Explanation:

There are theoretically three unique triangles that can be formed with two side lengths of 10 centimeters and one 40º angle. We include a representation of possible triangles:

Case A - Law of Cosine (
`a = 10\,cm`,
`b = 10\,cm`,
`C = 40^(\circ)`)

`c= \sqrt{a^(2)+b^(2)-2\cdot a\cdot b\cdot \cos C}` (1)

`c = \sqrt{(10\,cm)^(2)+(10\,cm)^(2)-2\cdot (10\,cm)\cdot (10\,cm)\cdot \cos 40^(\circ)}`

`c\approx 6.840\,cm`

Case B - Laws of Sine and Cosine (
`a = 10\,cm`,
`b = 10\,cm`,
`B = 40^(\circ)`)

`(a)/(\sin A) = (b)/(\sin B)` (2)

`\sin A = (a\cdot \sin B)/(b)`

`A = \sin^(-1)\left((a\cdot \sin B)/(b) \right)`

`A = \sin^(-1)\left[((10\,cm)\cdot (\sin 40^(\circ)))/(10\,cm) \right]`

`A = 40^(\circ)`

`C = 180^(\circ)-A-B`

`C = 100^(\circ)`

`c= \sqrt{a^(2)+b^(2)-2\cdot a\cdot b\cdot \cos C}`

`c = \sqrt{(10\,cm)^(2)+(10\,cm)^(2)-2\cdot (10\,cm)\cdot (10\,cm)\cdot \cos 100^(\circ)}`

`c \approx 15.321\,cm`

Case C - Laws of Sine and Cosine (
`a = 10\,cm`,
`b = 10\,cm`,
`A = 40^(\circ)`)

`(a)/(\sin A) = (b)/(\sin B)`

`\sin B = (b\cdot \sin A)/(a)`

`B = \sin^(-1)\left((b\cdot \sin A)/(a) \right)`

`B = 40^(\circ)`

`C = 180^(\circ)-A-B`

`C = 100^(\circ)`

`c= \sqrt{a^(2)+b^(2)-2\cdot a\cdot b\cdot \cos C}`

`c = \sqrt{(10\,cm)^(2)+(10\,cm)^(2)-2\cdot (10\,cm)\cdot (10\,cm)\cdot \cos 100^(\circ)}`

`c \approx 15.321\,cm`

There are two unique triangles that can be formed two side lengths of 10 centimeters and one 40º angle.