Question

In a right triangle ABC, Right angled at C in which AB=13cm,BC=5cm, determine the value of cos^2B+sin^2A. ​

Answer 1

$(5 \div 13)^2 + (5 \div 13)^2$

$2 \times 25 \div 169$

50/169

Answer 2

here

by Pythagoras theorem

AC = 12 cm

Now

cos B = BC / AB = 5/13

And

sin A = BC / AB

Therefore

cos²B + sin²A = (5/13)² + (5/13)²

= 25/169 + 25/169

= 50/169

Hope it will help you