Question

1.) Given that the perimeter of a triangle is 216. If the angles of the triangle are in the ratio 5:6:7, determine the longest side of the triangle.

2.) 4 times the sine of a certain angle is 6 times of the square of the cosine of the same angle. Find the angle in degrees if it’s in the first quadrant.

3.) Determine k so that the points A(3,8), B(-1,0), and C(k,-2) are the vertices of a right triangle with right angle at B.

Answer 1

that We know 5x + 6x + 7 x 216 :-) 18x 216 216 «12 -) 18 gy ( Answers 1 X3 7*12= The longool sede briangle the q 4 sino 6 reaso sino =) viosos als y sino = costo= y sinto a (1-сoso) 4 9 y cos²o 9 Coso= 9 y cos²o + coso- 4 9 17 64 -1 + 81 → Cos o= 002-4149)(-4) (14) -750:18125 -| E 0,42573466 0.88889 0.88889 A AB - Be= c (6,2) -1, 60395, -0.646048427 =) ( 130.244 330 (This is in and quadrant) (3,8) (8 0) + (3-(+)² = √64 + 16 = 580 @ +29"+ H-K) ? Ju +(x+1)? ACEJ (+2) +(3-6)} = V100 + (3-K72 theorem, By using pythagorean (-1,0) AB² + B C² = Ac2 80 + 4+ (x+1)? 100 + (3-6) 2 84 + k + 2K +1 = 100+ 96-6K + K3 = & K= 24 3 Hence, the value 8K = 24 8 kis 13.