Question

triangle ABC is right angled at c if p is the length of the perpendicular from c to a b and a, b, c are the length of the opposite sides of angle a angle b angle c respectively then prove that 1 by p square is equals to 1 by a square + 1 by b square

Answer 1

Answer:

Step-by-step explanation:

Ab=c

Ac=bq

Cb=a

Cd=p

In triangle ABC= 1/2×b×h

1/2×BA× CD

1/2×c×p (BA= c,CD=p)

1/2cp i

In triangle ABC

Area of triangle = 1/2×b×h

1/2×CB×Ac

1/2×a×b ii

Equating i and ii

cp=ab

In triangle ACB

Ab^2=Ac^2+Cb^2

C^2=b^2+a^2

ab^2÷p^2=b^2+a^2 (from proved first

1/p^2=b^2+a^2/a^2b^2

1/p^2=b^2/a^2b^2+a^2/a^2b^2

1/p^2=1/a^2+1/b^2

Hence proved