The problem above uses a combination of sine and cosine law of triangle to solve for the m∠B.
Given:
<span>m∠A = 60°, b = 9 and c = 6
Cosine Law:
a^2=b^2+c^2-2bccosA
a^2=(9)^2+(6)^2-2(9)(6)cos(60)
a^2=81+36-54
a^2= 63
a=</span>

Sine Law

sin B= 9/

sin 60
sin B= 0.98198
B= sin ^-1 (0.98198) = 79.10
Therefore, m∠B is equal to 79.10
Answer:
$910
Step-by-step explanation:
700 x 0.30 = 210
700 + 210 = $910
Hello,
1) Conditions:
x+2>=0==>x>-2
x>=0
So x>=0
1/5 ln(x+2)^5 +1/2 [ln x -ln(x^2+3x+2)^2]
=ln(x+2)+ ln√x - ln(x+1)(x+2)
The area of any quad = the base* height
for q 1:
the base=12 in
the height= 3 in
then the area = 12*3 = 36 sq in
for q 2:
the base= 14 in
the height = 4.5 in
then the area =4.5*14 =63
Answer: the simplify anwser is 64a^ 6 - 27b^6
Step-by-step explanation: