Question

Find distance (acosA+bsinA,0) and (0,asinA-bcosA)

Answer 1

Step-by-step explanation:

distance between two point A(x,y) & B(p,q)

is given by

=square root of [(x-p)²+(y-q)²]

=√[(x-p)²+(y-q)²]

Sony this expression

distance between (acosA+bsinA,0) & (0,asinA-bcosA)

is given by

=square root [( acosA+bsinA-0)²+(0-(asinA-bcosA))²]

=square root[( acosA+bsinA)²+(-asinA+bcosA)²]

=square root[( acosA+bsinA)²+(bcosA-asinA)²]

=square root[( a²cos²A+b²sin²A+2abcosAsinA)+(b²cos²A+a²sin²A-2abcosAsinA)]

by grouping terms

=square root[( a²cos²A+a²sin²A+2abcosAsinA)+(b²cos²A+b²sin²A-2abcosAsinA)]

=square root[( a²(cos²A+sin²A)+b²(cos²A+sin²A)]

=square root[( a²*1+b²*1]

=√(a²+b²)

Answer 2

Step-by-step explanation:

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