Question

if secA=3/4 calculate the value of sinA and cosA​

Answer 1

Answer:

secA= 3/4=H/B

B=4;H=3

P=

$\sqrt{3 { }^{2} } - 4{}^{2} = \sqrt{} 9 - 16= - \sqrt { 7}$

P=√-7

sinA= P/H= √-7/3

cosA=4/3