How do you solve the right triangle given $a=32m, A=46°$ ?

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Answer 1 · 2016-09-28T14:51:20

The angles are $90°, 46° and 44°$
The sides are $32m, 30.9m and 44.5m$

Let's call the triangle ABC, where $hatB$ is the 90° angle.

The sum of the angles in a triangle is 180°

If $hatB = 90°,$ then $hatA +hatC=90°$

From $hatA = 46°$ , you can calculate $hatC$

$hatC = 90°-46° = 44°$

Using trigonometry to find the length of $AB or c$

$"opp"/"adj" = c/32 = tan44°$

$c = 32tan44°$ 3

$c = 30.9m$

The length of the hypotenuse, AC, can be found by using trig or Pythagoras's Theorem.

Pythagoras $color(white)(xxxxxxxxxxxxxxxxxx)$ Trig

$a^2+ c^2 = b^2color(white)(xxxxxxxxcccccxxx xx)"a"/32 = 1/(cos44)$

$32^2 + 30.9^2 =1978.9color(white)(xxxxxxxxx)a = 32/(cos44)$

$sqrt1978.9 = 44.5mcolor(white)(xxxxxxxxxxx) a = 44.5m$

How do you solve the right triangle given a=32m, A=46° a=32m, A=46°?