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?

3 Answers
Binayaka C.
May 31, 2018

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]

Dean R.
May 31, 2018

#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}}#

Reactionary Green Negroid
May 31, 2018

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.

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