Question

If sides A and B of a triangle have lengths of 6 and 12 respectively, and the angle between them is `(pi)/8`, then what is the area of the triangle?

Answer 1

Area of the triangle is `13.78` sq.unit.

Explanation:

Angle between Sides `A and B` is

`/_c= pi/8=180/8=22.5^0`. Sides are `A=6 , B=12`

Area of the triangle is `A_t=(A*B*sin c)/2` or

`A_t=(6*12*sin 22.5)/2~~ 13.78` sq.unit

Area of the triangle is `13.78` sq.unit [Ans]

Answer 2

`text{area} = 1/ 2 a b sin C = 1/2 (6)(12) sin (pi/8) = 18 sqrt{2-sqrt{2}}`

Explanation:

`sin(pi/8) = sin(45^circ/2) = sqrt{1/2(1 - cos 45^circ)} =sqrt(1/2(1-sqrt{2}/2)) = 1/2 sqrt{2 - sqrt{2}}`

Answer 3

13.8

Explanation:

Let's plot this triangle on an imaginary two-dimensional graph so that:

Your given angle (we can call it `theta`) is located at the origin, with line A on the `x` axis and line B slanting upwards and rightwards at the angle `pi`/`8` from the origin.

We can now calculate the y-component of line B as `12*sin(theta)` which is approximately 4.6 units.

We can now multiply the 'base' of the triangle (A) by the 'height' of the triangle (`b_y`) and we get `6*4.6 = 27.6`. Halve that and you get your area.