Question

How do you work out this problem

Answer 1

Answer:

Sin A = $\frac{2}{3.6}$

Cos A =$\frac{3}{3.6}$

Tan A =$\frac{2}{3}$

Step-by-step explanation:

First of all, we will need to find the hypotenuse. We can do this using the Pythagoras theorem.

$c^{2} = a^{2} + b^{2}$

where a = 2 and b = 3,

solving for c we end up with 3.6.

Sin A = $\frac{opp}{hyp}$

Cos A = $\frac{adj}{hyp}$

Tan A = $\frac{opp}{hyp}$

The line opposite to the angle A is 2 and the line adjacent to it is 3.

Filling in, we get the answer.

Hope this helps.

Answer 2

Answer:

$sin(A) = \frac{2}{\sqrt{13} }=\frac{2\sqrt{13} }{13}$  

$cos(A)=\frac{3}{\sqrt{13} } =\frac{3\sqrt{13}}{13}$

$tan(A)=\frac{2}{3}$

Step-by-step explanation:

So first we need to make sure we know the trig identities to solve this problem.

• $sin(\alpha )=\frac{opposite}{hypotenuse}$
• $cos(\alpha )=\frac{adjacent}{hypotenuse}$
• $tan(\alpha )=\frac{opposite}{adjacent}$

Here we only have two legs of the triangle, so we will need to use the Pythagorean Theorem a² + b² = c² to solve for the missing leg, the hypotenuse in this case.

• Solving for the hypotenuse, c, we get $c = \sqrt{a^{2} +b^{2} }$
• Here a = 3 and b = 2, so plugging in these values to the equation we get: $c= \sqrt{(3)^{2}+(2)^{2} } =\sqrt{9+4} =\sqrt{13}$

Now we can use the trig identities to figure out the missing values for the problem

• $sin(A) = \frac{2}{\sqrt{13} }=\frac{2\sqrt{13} }{13}$
• $cos(A)=\frac{3}{\sqrt{13} } =\frac{3\sqrt{13}}{13}$
• $tan(A)=\frac{2}{3}$