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In triangle ABC, how do you solve the right triangle given 'a' is 7 m, and the angle 'A' is 22 degrees?

1 Answer

Sep 29, 2015

(Assuming the right angle is at C)
$∠ B = 68^{\circ}$ $/_B = 68^@$
$c = \frac{7}{sin (22^{\circ})} m e t e r s$ $c= 7/sin(22^@) meters$
$b = \frac{7}{tan (22^{\circ})} m e t e r s$ $b=7/tan(22^@) meters$

Explanation:

Everything that follows is under the assumption that the right angle is $∠ C$ $/_C$ . If the intent is that $∠ B$ $/_B$ is the right angle, exchange $B$ $B$ and $C$ $C$ in the explanation below:

Since $∠ A = 22^{\circ}$ $/_A =22^@$ and $∠ C = 90^{\circ}$ $/_C = 90^@$
and the sum of interior angles of a triangle is $180^{\circ}$ $180^@$
$\to ∠ B = 68^{\circ}$ $rarr /_B = 68^@$

For the $∠ A$ $/_A$
the opposite side is $a (= 7)$ $a (=7)$ m
the adjacent side is $b$ $b$ and
the hypotenuse is $c$ $c$

$sin (22^{\circ}) = \frac{opposite}{hypotenuse} = \frac{7}{c}$ $sin(22^@) = ("opposite")/("hypotenuse")= 7/c$

$\to c = \frac{7}{sin (22^{\circ})}$ $rarr c= 7/sin(22^@)$

and

$tan (22^{\circ}) = \frac{opposite}{adjacent} = \frac{7}{b}$ $tan(22^@) = ("opposite")/("adjacent") = 7/b$

$\to b = \frac{7}{tan (22^{\circ})}$ $rarr b=7/tan(22^@)$

Note:
a calculator could be used to evaluate:
$XXX b = \frac{7}{tan (22^{\circ})} ≅ 17.33$ $color(white)("XXX")b=7/tan(22^@) ~=17.33$
and
$XXX c = \frac{7}{sin (22^{\circ})} ≅ 18.69$ $color(white)("XXX")c=7/sin(22^@) ~=18.69$

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