Question

Pls solve no;3a,b,c the picture will begiven thank u​ use trigonometry u solve it

Answer 1

Answer/Step-by-step explanation:

3.a. length of AC.

Hypotenuse = AC

Opposite side to 60° = 7

Therefore, $sin(60) = \frac{7}{AC}$

$AC*sin(60) = \frac{7}{AC}*AC$

$AC*sin(60) = 7$

$AC = \frac{7}{sin(60)}$

$AC = 8.1$ (to nearest tenth)

b. Length of BC

Hypotenuse = BC

Opposite side to 50° = 7

Therefore, $sin(50) = \frac{7}{BC}$

$BC*sin(50) = \frac{7}{BC}*AC$

$BC*sin(50) = 7$

$BC = \frac{7}{sin(50)}$

$BC = 9.1$ (to nearest tenth)

c. Length of AB

Use the law of sines to find AB

$\frac{BC}{sin(A)} = \frac{AB}{sin(C}}$

m<C = 180 - (60 + 50) (sum of angles in a triangle) = 70°

m<A = 60°

BC = 9.1

AB = ?

$\frac{9.1}{sin(60)} = \frac{AB}{sin(70)}$

Cross multiply

$9.1*sin(70) = AB*sin(60)$

Divide both sides by sin(60)

$\frac{9.1*sin(70)}{sin(60)} = \frac{AB*sin(60)}{sin(60)}$

$\frac{9.1*sin(70)}{sin(60)} = AB$

$9.9 = AB$ (nearest tenth)

$AB = 9.9$