# The right angled triangle ABC has AB=6, BC=8 and AC is the hypotenuse. Find sinA, cosA and tanA.

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### 3 Answers

We have the right triangle ABC where AC is the hypotenuse and AB= 6 and BC = 8.

By the Pythagorean theorem: AC^2 = AB^2 + BC^2

=> AC^2 = 6^2 + 8^2

=> AC = sqrt [ 36 + 64]

=> AC = sqrt 100

=> AC = 10

Now **sin A = 8/10 = 4/5**

**cos A = 6/10 = 3/5**

and **tan A = 8/6 = 4/3**

Given the the right angle triangle ABC such that AC is the hypotenuse.

Then, the right angle is B.

==> Given the legs are:

AB = 6

BC = 8

Then, we will calculate the hypotenuse AC using the formula.

==> AC^2 = BC^2 + AB^2

==> AC^2 = 6^2 + 8^2 = 36+64 = 100

==> AC = 10

Now we need to find the following:

sinA = opposite/ hypotenuse.

We know that the side that is opposite to the angle A is BC

==> sinA = BC/AC = 8/10 = 4/5

==> cosA = adjacent/ hypotenuse

= AB/AC = 6/10 = 3/5

==> tanA = sinA/cosA = 4/3

**==> sinA = 0.8 **

**==>cosA = 0.6**

**==> tanA = 4/3**

In the right angled triangle ABC, AB = 6 , BC = 8 .

So we calculate hypotenuse AC = {(AB)^2+(BC^2)}^(1/2) in accordance with the Pythagoras theorem.

=> AC = (6^2+8^2)^(1/2) = (36+64)^(1/2) = 100^(1/2) = 10.

Therefore SinA = Opposite side of angle A /Hypotenuse = BC/AC = 8/10 = 0.8.

CosA = Adjacent side of angle A / Hypotenuse = AB/AC = 6/10 = 0.6.

TanA = Opposite side of angle A /Adjacent side of angle A = 8/6 = 4/3.