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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))

Sagot :

Medunno13idn

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

[tex]x+2x+\sqrt{x^{2}+(2x)^{2}}=28\\3x+x\sqrt{5}=28\\x(3+\sqrt{5})=28\\x=\frac{28}{3+\sqrt{5}} \cdot \frac{3-\sqrt{5}}{3-\sqrt{5}}=21-7\sqrt{5}[/tex]

This means that the length of the hypotenuse is:

[tex]\sqrt{5}(21-7\sqrt{5})=\boxed{21\sqrt{5}-35}[/tex]

2) (Diagram attached) We know that

[tex]\cos 51^{\circ}=\frac{x}{58}\\x=58 \cos 51^{\circ} \approx \boxed{36.5}[/tex]  (in cm)

3) From the Pythagorean identity,

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

4) [tex]\frac{\pi}{6}[/tex]

View image Medunno13
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