Question

Triangle ABC has an area of 25√3/2 cm²,c=60•, a=5cm
find the angle of A and B​

Answer 1

The possible solutions for the triangle ABC are:

(
`a = 5\,cm`,
`b = 64.952\,cm` and
`c = 60\,cm`):

`A = 0.634^(\circ)`,
`B = 171.731^(\circ)`

(
`a = 5\,cm`,
`b = 55.057\,cm` and
`c = 60\,cm`)

`A = 0.751^(\circ)`,
`B = 8.293^(\circ)`

Explanation:

The area of the triangle ABC is determined by the following equation:

`A = √(s\cdot (s-a)\cdot (s-b)\cdot (s-c))`

`s = (a+b+c)/(2)`

Where:

`A` - Area of the triangle, measured in square centimeters.

`s` - Semiperimeter of the triangle, measured in centimeters.

`a`,
`b`,
`c` - Sides of the triangle, measured in centimeters.

Now, we simplify the equation:

`A^(2) = s\cdot (s-a)\cdot (s-b)\cdot (s-c)`

`A^(2) = \left((a+b+c)/(2)\right)\cdot \left((b+c-a)/(2) \right)\cdot \left((a+c-b)/(2) \right)\cdot \left((a+b-c)/(2) \right)`

`16\cdot A^(2) = (a+b+c)\cdot (b+c-a)\cdot (a+c-b)\cdot (a+b-c)`

If we know that
`A = 25\cdot (√(3))/(2)\,cm^(2)`,
`a = 5\,cm` and
`c = 60\,cm`, the equation is simplified:

`7500 = (b+65\,cm)\cdot (b+55\,cm)\cdot (65\,cm-b)\cdot (b -55\,cm)`

`7500\,cm^(2) = -(b^(2)-4225\,cm^(2))\cdot (b^(2)-3025\,cm^(2))`

`7500\,cm^(2) = -(b^(4)-7250\cdot b^(2) +12780625\,cm^(2))`

`7500\,cm^(2) = -b^(4)+7250\cdot b^(2) -12780625\,cm^(2)`

`b^(4)-7250\cdot b^(2)+12788125\cm^(2) = 0`

The roots of the fourth-grade polynomial are:

`b_(1) \approx 64.952\,cm`,
`b_(2) \approx 55.057\,cm`,
`b_(3)\approx -55.057\,cm`,
`b_(4)\approx -64.952\,cm`.

Only the first two roots are physically reasonable. Then, there are only two solutions for the triangle ABC:

`b_(1) \approx 64.952\,cm`,
`b_(2) \approx 55.057\,cm`

The angles A and B can be found by Law of Cosine:

`a^(2) =b^(2)+c^(2)-2\cdot b\cdot c \cdot \cos A`

`2\cdot b\cdot c\cdot \cos A = b^(2)+c^(2)-a^(2)`

`\cos A = (b^(2)+c^(2)-a^(2))/(2\cdot b\cdot c)`

`A = \cos^(-1)\left((b^(2)+c^(2)-a^(2))/(2\cdot b\cdot c) \right)`

`b^(2) = a^(2)+c^(2)-2\cdot a\cdot c \cdot \cos B`

`2\cdot a \cdot c\cdot \cos B = a^(2)+c^(2)-b^(2)`

`\cos B = (a^(2)+c^(2)-b^(2))/(2\cdot a\cdot c)`

`B =\cos^(-1)\left((a^(2)+c^(2)-b^(2))/(2\cdot a \cdot c) \right)`

Where:

`A`,
`B` - Angles opossed to sides
`a` and
`b`, measured in sexagesimal degrees.

There are two possibilities:

(
`a = 5\,cm`,
`b = 64.952\,cm` and
`c = 60\,cm`)

`A = \cos^(-1)\left[((64.952\,cm)^(2)+(60\,cm)^(2)-(5\,cm)^(2))/(2\cdot (64.952\,cm)\cdot (60\,cm)) \right]`

`A = 0.634^(\circ)`

`B = \cos^(-1)\left[((5\,cm)^(2)+(60\,cm)^(2)-(64.952\,cm)^(2))/(2\cdot (5\,cm)\cdot (60\,cm)) \right]`

`B = 171.731^(\circ)`

(
`a = 5\,cm`,
`b = 55.057\,cm` and
`c = 60\,cm`)

`A = \cos^(-1)\left[((55.057\,cm)^(2)+(60\,cm)^(2)-(5\,cm)^(2))/(2\cdot (55.057\,cm)\cdot (60\,cm)) \right]`

`A = 0.751^(\circ)`

`B = \cos^(-1)\left[((5\,cm)^(2)+(60\,cm)^(2)-(55.057\,cm)^(2))/(2\cdot (5\,cm)\cdot (60\,cm)) \right]`

`B = 8.293^(\circ)`

The possible solutions for the triangle ABC are:

(
`a = 5\,cm`,
`b = 64.952\,cm` and
`c = 60\,cm`):

`A = 0.634^(\circ)`,
`B = 171.731^(\circ)`

(
`a = 5\,cm`,
`b = 55.057\,cm` and
`c = 60\,cm`)

`A = 0.751^(\circ)`,
`B = 8.293^(\circ)`