+0

Two Trig Question Help Plz

-1
143
1
+34

1.

2.

Jun 5, 2021

#1
+179
+1

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!

Jun 5, 2021