in any triangle a
let a, b, c be the sides of any triangle abc. then by applying the sine rule, we get⇒c = k sin csimilarly, b = k sin band a = k sin ahere we will consider rhs, so we getrhs = (b2 – c2) sin asubstituting corresponding values in the above equation, we get⇒ = [( k sin b)2 - ( k sin c)2] sin a⇒ = k2(sin2 b – sin2 c )sin a……….(ii)but,substituting the above values in equation (ii), we get⇒ = k2(sin(b + c) sin(b - c)) sin abut a + b + c = π ⇒ b + c = π –a, so the above equation becomes,⇒ = k2(sin(π –a) sin(b - c))sin abut sin (π - θ) = sin θ⇒ = k2(sin(a) sin(b - c))sin arearranging the above equation we get⇒ = (k sin(a))( sin(b - c))(k sin a)from sine rule, a = k sin a, so the above equation becomes,⇒ = a2 sin(b - c) = rhshence proved
a, b and c a - ii, b - iv, c - i, d - iii a, b & c 3 a
c
c
a
c, i think please the first one. hello, red.since the triangle is a right triangle, you can use the pythagorean theorem.let x be the length of the unknown side.then,92 + x2 = (9√2)281 + x2 = 81·2 = 162x2 = 162 - 81x2 = 81take the square root of both sides.x = ±9since the length cannot be a negative number, x = 9 m dear student,
i b mcleery
ii c hastily
iii d rubber ring
iv a to ensure that he had no objectionable material with him
regards Notice that the matrix A is upper-triangle, so its determinant is a  
product   
of diagonal elements:  
  
det(A) = 1*2*3 = 6.  
  
Moreover, det(A) is   
non-zero.  
Therefore the columns of A are linearly independent and   
therefore  
they constitutive a basis for R^3, so R^3 = Col(A).  
  
Hence   
any base of R^3 fits to be a base for Col(A),   
for instance the columns of   
A:  
  
(1,0,0), (2,2,0), (3,3,3)  
  
or   
  
(1,0,0), (0,1,0),   
(0,0,1)  
  
are the base.  
  
The dimension of Col(A) is the dimension of   
R^3 and therefore is equal  
to 3.