Answer:
13
Step-by-step explanation:
Diagonals of a rhombus are perpendicular bisector.
Hence, triangle ABE so formed would be a right triangle right angled at E.
Therefore, by Pythagoras theorem:
(3 cos x-4 sin x)+(3sin x+4 cos x)=5
(3cos x+4cos x)+(-4sin x+3 sin x)=5
7 cos x-sin x=5
7cos x=5+sin x
(7 cos x)²=(5+sinx)²
49 cos²x=25+10 sinx+sin²x
49(1-sin²x)=25+10 sinx+sin²x
49-49sin²x=25+10sinx+sin²x
50 sin² x+10sinx-24=0
Sin x=[-10⁺₋√(100+4800)]/100=(-10⁺₋70)/100
We have two possible solutions:
sinx =(-10-70)/100=-0.8
x=sin⁻¹ (-0.8)=-53.13º (360º-53.13º=306.87)
sinx=(-10+70)/100=0.6
x=sin⁻¹ 0.6=36.87º
The solutions when 0≤x≤360º are: 36.87º and 306.87º.
Answer: Yes
Step-by-step explanation:
f=3
11+3= 14
23 > 14
es correcto decir que 23 es mas grande que 11+f
Answer:
AD=5.82 units
Step-by-step explanation:
tan 31=3/BD
0.6009=3/BD
BD=5
AD^2=3^2 + 5^2
AD^2=9+25=34
AD=5.82
