How do you solve the triangle if it has legs with lengths of 18 and 28 and there is a 90 degree angle, X and Y are the other two angles?

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Answer 1 · 2015-09-30T18:43:18

Use Pythagorean theorem to find the length of the hypotenuse then use trigonometry to solve for the missing angles.

Below is a sketch of the triangle XYZ:

enter image source here

Using Pythagorean Theorem:

${(¯ ¯¯¯¯ ¯ X Y)}^{2} = {(¯ ¯¯¯¯ ¯ X Z)}^{2} + {(¯ ¯¯¯¯ ¯ Y Z)}^{2}$ $(bar(XY))^2 = (bar(XZ))^2+(bar(YZ))^2$

$¯ ¯¯¯¯ ¯ X Y = \sqrt{1108} = 33.3$ $bar(XY)= sqrt(1108) = 33.3$

Now, use the property of inverse trig function (inverse tangent) to solve for angle Y:

$tan Y = \frac{18}{28}$ $tan Y = 18/28$

$∠ Y$ $/_Y$ = ${32.7}^{o}$ $32.7^o$

Finally, $∠ X = 180^{o} - 90^{o} - {32.7}^{o} = {57.3}^{o}$ $/_X = 180^o-90^o-32.7^o=57.3^o$

Hope that helped