We know that, in ∆ABC,
∠A+∠B+∠C = 180°
But the triangle is right angled at C
ie., ∠C = 90°
Therefore, ∠A+∠B+ 90° = 180°
⇒ ∠A + ∠B = 90°
Therefore, <u>cos(A + B) = cos 90º = 0</u>
Answer:
x^2 + (y+1)^2 = 10^2
Step-by-step explanation:
Centre (p,q) = (0, -1)
radius, r = 10
General Equation of a circle
(x-p)^2 + (y-q)^2 = r^2
(x-0)^2 + (y- -1)^2 = 10^2
x^2 + (y+1)^2 = 10^2
sin 45 = opp/hyp
sin 45 = 4 / y
y = 4/sin 45
y = 4 /((sqrt2)/2)=4* sqrt(2)
tan 45 =opp/adj
tan 45 = 4/(x-3)
1 = 4/(x-3)
x-3 = 4
x=7
B
Answer:
x= 5 1/30
Step-by-step explanation:
hope this is fine :)
My Answer: Pentagons( First image)
Hope I helped! :D
