Question

if sin A = 3/4 then calculate cos A and tan A
why we Cant take AB=3k,BC=4k​

Answer 1

Answer:

if sin theta +cos theta =k then 2(sin^3 theta+cos^3 theta )=

a) 3k-4k^3

b)2k+4k^3

c)3k+k^3

d)3k-k^3

Answer 2

Answer :-

Given ,

• sinA = 3/4

So ,

• Opposite side of angle A = 3
• Hypotenuse of triangle = 4

The third side ,

By Pythagoras theorem ,

Hypotenuse² = Base² + height²

Base² = Hypotenuse² - Height²

Base = √ (3)² + (4)²

Base = √9+16

Base = √25

Base = 5

Now ,

• CosA = Adjacent (Base)/hypotenuse = 5/4
• TanA = Opposite/Adjacent = 3/5

Hence , tanA = 3/5 & cosA = 5/4

We can take it as 3k & 4k because , it will be sum of their square and rooted i.e √(3k)²+(4k)² = √25k² = √(5k)² = 5k . But again the sine value should be multiplied and divided by k , ↓

Now , sinA = 3k/4k

Base = 5k

cosA = 5k/4k = 5/4

tanA = 3k/5k = 3/5

We get the same answer in any way !.