Question

18th sum solved answer ​

Answer 1

Answer:

what's the question first

Answer 2

Answer:

1.Draw CE perpendicular to AD

In triangle ABC, Right angle at B

By pythagoras theorem

AB^2+BC^2=AC^2

AC^2-BC^2=AB^2

13*13-5*5=AB^2

AB^2=144

AB=12

COS Teeta =12/13

In triangle AEC, AE=7units, in triangleDCE, DC=13

DE=7

By pythagoras theorem

13*13-7*7=CE^2

CE^2=120

CE=

$2 \sqrt{30}$

2.Tan phi=2 *square root 30/7

3.Cosec phi=2*square root 30/13