How do you solve the triangle if it has hypotenuse 18, the opposite angle is the height, and the adjacent meets the hypotenuse at a 35 degree angle?

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How do you solve the triangle if it has hypotenuse 18, the opposite angle is the height, and the adjacent meets the hypotenuse at a 35 degree angle?

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Answer 1 · 2015-06-08T17:42:09

Assuming you meant "the opposite side is the height"

If we label the triangle $A B C$ $ABC$ with
$∠ A = 35^{\circ}$ $/_A = 35^@$
and
$∠ B = 90^{\circ}$ $/_B=90^@$

then
$∠ C = 55^{\circ}$ $/_C = 55^@$

Given $c = 18$ $c=18$ ( $c$ $c$ being the side opposite $∠ C$ $/_C$ )

then we can use the Law of Sines:
$XXXX$ $color(white)("XXXX")$ $\frac{sin C}{c} = \frac{sin A}{a} = \frac{sin B}{b}$ $(sin C)/c = (sin A)/a = (sin B)/b$
and since our only missing values are $a$ $a$ and $b$ $b$
the results can easily be calculated (with a calculator or with spreadsheet operations)

Answer 2 · 2015-06-08T18:02:52

C = 90; B = 35; and A = 90 - 35 = 55 deg; hypotenuse = 18

Side CB = AB.cos 35 = 18.cos 35 = 18(0.82) = 14.75

The height, or side AC = 18.sin 35 = 18(0.57) = 10.32

Check: a^2 = b^2 + c^2

324 = 217.56 + 106.50 = 324.06 . OK

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How do you solve the triangle if it has hypotenuse 18, the opposite angle is the height, and the adjacent meets the hypotenuse at a 35 degree angle?

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