Question

in triangle abc ab=ac=15cm bc=18 find tan^2b-sec^2b+2​

Answer 1

Answer:

AB=AC=15, BC=18

Using cosine rule,

AC2=BC2+AB2−2(AB)(BC)cosB

152=182+152−2(15)(18)cosB

cosB=159

cosB=53

cosB=HB=53

Now, using Pythagoras theorem.

H2=P2+B2

52=P2+32

P=4

tan2B−sec2B+2

= (BP)2−(BH)2+2

= (34)2−(35