Question 1199591
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I'm a triangle ABC.
Cos A/a = Cos B/b = Cos C/c, and a = 2. It is possible to compute the area of the
triangle, (Hint: compare the problem data with the Law of the Sines, compute ratios between the sines and cosines of the angles)
A) sqrt3/4
B) sqrt3/2
C) sqrt3
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<pre>
We are given  Cos(A)/a = Cos(B)/b = Cos(C)/c.    (1)


Let "p" be the common value of all these three ratios

    {{{cos(A)/a}}} = {{{cos(B)/b}}} = {{{cos(C)/c}}} = p.    (2)

Then

    cos(A) = pa,    (3)

    cos(B) = pb,    (4)

    cos(C) = pc.    (5)


From the other side, we have this "Law of Sines" formula

    {{{sin(A)/a}}} = {{{sin(B)/b}}} = {{{sin(C)/c}}}.    (6)


Let "q" be the common value of all these three ratios

    {{{sin(A)/a}}} = {{{sin(B)/b}}} = {{{sin(C)/c}}} = q.    (7)

Then

    sin(A) = qa,    (8)

    sin(B) = qb,    (9)

    sin(C) = qc.    (10)


Divide (8) by (3);  divide (9) by (4);  divide (10) by (5).  You will get then

    {{{sin(A)/cos(A)}}} = {{{q/p}}},

    {{{sin(B)/cos(B)}}} = {{{q/p}}},

    {{{sin(C)/cos(C)}}} = {{{q/p}}},


or  tan(A) = tan(B) = tan(C) = {{{q/p}}}.


The tangents of three angles of a triangle are equal --- hence the angles are congruent A = B = C.


So, the triangle ABC is an equilateral triangle with the side length of 2 units.


Hence, its area is  {{{a^2*(sqrt(3)/4)}}} = {{{4*(sqrt(3)/4)}}} = {{{sqrt(3)}}} square units.
</pre>

Solved.