Question

if 5cosA=4 and angleA is acute then find tanA
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Answer 1

Answer:

then SinA= 4/5, CosA= 3/5 where A is an acute angle of right triangle . Now, = \frac{\sqrt{1 - sinA}}{{1+CosA.

Answer 2

Step-by-step explanation:

Tan A = 4/3,

As, tan A= perpendicular/Base,

By pythagoras theorem

(perpendicular)² + (Base)² =( hypotenuse)²

4² + 3²=( hypotenuse)²

25=( hypotenuse)²

Hypotenuse=5 units

then SinA= 4/5, CosA= 3/5 where A is an acute angle of right triangle .

Now, =\frac{\sqrt{1 - sinA}}{{1+CosA}} =\frac{\sqrt{1-\frac{4}{5}} }{1+ \frac{3}{5}}=\frac{\frac{1}{\sqrt5}} {\frac{8}{5}}=\frac{5}{8\sqrt5}=\frac{\sqrt5}{8}

1+CosA

1−sinA