Question

plz solve this
I will mark you as brainliest​

Answer 1

Answer:

$\sin(a) = \frac{3}{4} = \frac{p}{h}$

by pythagores theorem

${h}^{2} = {p}^{2} + {b}^{2}$

(4)^2=(3)^2+(b)^2

16=9+b2

${b}^{2} = 16 - 9 = 7$

$b = \sqrt{7}$

$\cos( a) = \frac{b}{h} = \frac{ \sqrt{7} }{4}$

$\tan(a) = \frac{p}{b} = \frac{3}{ \sqrt{7} }$

$\cot(a) = \frac{b}{p} = \frac{ \sqrt{7} }{3}$

$\sec(a) = \frac{h}{b} = \frac{4}{ \sqrt{7} }$

Answer 2

Answer:

I'm fine

why about you

Step-by-step explanation:

Mark upper one as brainliest