+179 1. ABC is an isosceles triangle, so the altitude from C to AB will divide the triangle into two identical right triangles, and be an angle bisector of the angle C. Let the point of intersection = D. Due to the fact that the sum of angles in a triangle is 180 degrees, and that this triangle is isosceles, angle a is 67.5 degrees.
Sin (A) = CD/3 = Sin (67.5). Thus, CD = 3Sin(67.5).
Cos (67.5) = AD/3. Thus, AD = 3Cos (67.5).
AB = 2AD = 6 Cos (67.5)
Area = AB*CD/2 = 3Sin(67.5)*6Cos(67.5)/2 = 9Sin(67.5)Cos(67.5) = 3.18198051534
2. Let AB = a and BC = b
Thus, Tan (c) = 2.4 translates to a/b = 2.4, or b = 1/2.4a
Cos a = a/AC = \(\frac{a}{\sqrt{a^2+b^2}}\)
Substituting b = 1/2.4a into the expression, we have Cos A = \(\frac{a}{\sqrt{a^2+\frac{1}{2.4^2}a^2}}\)
Cos A = \(\frac{a}{\sqrt{\frac{169}{144}a^2}}\)
=\(\frac{a}{\frac{13}{12}a}\)
=\(12/13 \)
If by chance you have memorized common triangle trigonometric relationships (I haven't myself), then recognizing that ABC is a
5-12-13 triangle right away would be a very quick method.
Long response, but I hope this answered your question. Have a great day!
