Question

One of the legs of a right triangle is twice as long as the other, and the perimeter of the triangle is 28. Find the length of the hypotenuse.

In a right triangle, the 58-cm hypotenuse makes a 51-degree angle with one of the legs. To the nearesttenth of a cm, how long is that leg?

Evaluate: sin (Arccos(5/13))

Evaluate: tan(Arctan(π/6))

Answer 1

1) If we let the lengths of the legs be x and 2x, then by the Pythagorean theorem,

`x+2x+\sqrt{x^(2)+(2x)^(2)}=28\\3x+x√(5)=28\\x(3+√(5))=28\\x=(28)/(3+√(5)) \cdot (3-√(5))/(3-√(5))=21-7√(5)`

This means that the length of the hypotenuse is:

`√(5)(21-7√(5))=\boxed{21√(5)-35}`

2) (Diagram attached) We know that

`\cos 51^(\circ)=(x)/(58)\\x=58 \cos 51^(\circ) \approx \boxed{36.5}` (in cm)

3) From the Pythagorean identity,

`\sin^(2)\left (\arccos (5)/(13) \right)+\cos^(2) \left (\arccos (5)/(13) \right)=1\\\left((5)/(13) \right)^(2)+\sin^(2) \left (\arccos (5)/(13) \right)=1\\(25)/(169)+\sin^(2) \left (\arccos (5)/(13) \right)=1\\\sin^(2) \left (\arccos (5)/(13) \right)=(144)/(169)\\\sin \left (\arccos (5)/(13) \right)=\boxed{(12)/(13)}`

4)
`(\pi)/(6)`