Question

(b) In the figure (2) given below, AABC is right angled at B and BD is perpendicular to
AC. Find
(i) cos ZCBD (ii) cot ZABD.​

Answer 1

Answer:

in ∆ABC,

by Pythagoras theorem,

AC^2=AB^2+BC^2

=(12)^2+(5)^2

=144+25

=169

hence,AC=√169=13

Hence,AC=13

Now,

as BD perpendicular AC

therefore,DC=1/2×AC

=13/2

Now,in ∆BCD,

(BC)^2 = (BD)^2+(13/2)^2

25=(BD)^2+169/4

BD=42.25-25

=17.25

Now,

CD/BD=cosCBD

cosCBD=( 13/2)÷17.25=6.5/17.25

cotABD=17.25/12