Use the given information to solve the triangle for A = 150°, b = 4.8, a = 9.4

1 answer:

7 0

Well you can start by drawing a triangle with the information

the size of angle B can be found using the Law of Sine

$\frac{sin(150)}{9.4)} = \frac{sin(B)}{4.8}$

the size of angle C can be found using angle sum of a triangle.

The length of side c can be found using the Law of Cosines

$c^2 = 4.8^2 + 9.4^2 - 2 \times4.8\times9.4\times cos(C)$

hope it helps

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Given:
A has coordinates (-6,5)
C has coordinates (3.6,-0.4).
C divides AB in the ratio 3:2.

Refer to the diagram shown below.
The coordinates of B are determined by
$\frac{3.6-(-6)}{x-(-6)} = \frac{3}{3+2} \\ \frac{9.6}{x+6} = \frac{3}{5} \\ 3(x+6) = 48 \\ 3x +18=48 \\ x=30/3=10$
Also,
$\frac{5-(-0.4)}{5-y} = \frac{3}{5} \\ 3(5-y) = 27 \\ 15-3y=27 \\ y= \frac{12}{-3} =-4$

Answer:
The coordinates of B are (10,-4).

Now, we know the coordinates of line segment AB as A (-6, 5) and B (10,-4).
D (x,y) divides AB in the ratio 4:5.
Therefore
$\frac{x-(-6)}{10-(-6)} = \frac{4}{4+5} \\ \frac{x+6}{16} = \frac{4}{9} \\ 9(x+6)=64 \\ 9x+54=64 \\ x=1.11$
Also,
$\frac{5-y}{5-(-4)} = \frac{4}{9} \\ 5-y = 4 \\ y = 1$

Answer:
The coordinates of D are (1.11, 1)