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You use the Pythagorian Theorum, a2 + b2 = c2, where c is the length of the hypotenuse.

If a2 + b2 = c2 then a2 = c2 - b2.

Let's say that b is the side opposite Î˜. Since sinÎ˜ = b/c, a2 = 5 - 4 = 1, so a = 1.

Since tanÎ˜ = b/a, tanÎ˜ = 2.

Q: How do you find tan theta when sin theta equals 2 over square root 5?

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You shoot a cannon with a vi of 18 m/s. You need to get it 16m and over a 10m wall. The acceleration in the x is 0 and in the y it is gravity. The question is at what angle theta can you shoot the cannon over the fence and 16 meters away. How do I find theta?

Ut is equual to tan(theta) / (sec(theta) + 1)

For such simplifications, it is usually convenient to convert any trigonometric function that is not sine or cosine, into sine or cosine. In this case, you have: sin theta / sec theta = sin theta / (1/cos theta) = sin theta cos theta.

There is none because it is not true.

area equals pi r squared therefor r squared equals area over pi. Now find square root of r squared and you have "R" (radius) = 2.821

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If sin (theta) is 3/5, then sin2 (theta) is (3/5)2, or 9/25.

You shoot a cannon with a vi of 18 m/s. You need to get it 16m and over a 10m wall. The acceleration in the x is 0 and in the y it is gravity. The question is at what angle theta can you shoot the cannon over the fence and 16 meters away. How do I find theta?

zero

Ut is equual to tan(theta) / (sec(theta) + 1)

For such simplifications, it is usually convenient to convert any trigonometric function that is not sine or cosine, into sine or cosine. In this case, you have: sin theta / sec theta = sin theta / (1/cos theta) = sin theta cos theta.

There is none because it is not true.

area equals pi r squared therefor r squared equals area over pi. Now find square root of r squared and you have "R" (radius) = 2.821

Assuming you know the angle of ascension, and the base, you can calculate the height by recalling that tangent theta is height over base. Simple algebra from there: height is tangent theta times base.

In a right triangle, the right angle is formed by sides a and b. Side c is the hypotenuse.Theta is the interior angle that joins (let's say) sides b and c. The sin of theta is the length of a over the length of c. The cos of theta is the length of b over the length of c. The tan of theta is the length of a over the length of b.Sin theta= opposite divided by hypotenuse. Cos theta=adjacent divided by hypotenuse. Tan theta=opposite over adjacent.(Sin1-Cos1)-Tan1=-1.25623905Sorry, that was a mathematician's joke.

If you move that 46x over the equals, and if you square both sides, you can get rid of that square root, and do the equation normally. do the math

Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta

The square root of 25/36 equals 5/6