Question

1) in a ∆ ABC, LB =90° and AB=4 cm, BC=3cm and BC=5cm. find i)tanA ii) sin^2A + cos^2A​ . with solution​

Answer 1

1• tan A

= P/ b

=3/4

2• sin2A +cos 2A

= 2× p/h + 2× b/h

= 2× 3/5 + 2× 4/5

=6/5 + 8/5

= 14/5

Here, p= perpendicular , b for base and h for hypotenuse.

Hope this help u

Thanku

Answer 2

Answer:

Answer is 14/5

Step-by-step explanation:

1. Tan A

=p/b

=3/4

2. Sin 2A+Cos 2A

=2*p/h+2*b/h

=2*3/5+2*4/5

6/5+8/5

=14/5

P=perpendicular

B=Base

H=Hypotenuse