How do you find the measure of each of the angles of a triangle given the measurements of the sides are 32, 40, 38?

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How do you find the measure of each of the angles of a triangle given the measurements of the sides are 32, 40, 38?

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Answer 1 · 2016-12-31T11:11:28

The angles are $∠ A = {48.36}^{0}, . ∠ B = {69.09}^{0}, ∠ C = {62.55}^{0}$ $/_A=48.36^0, ./_B=69.09^0,/_C= 62.55^0$

Explanation:

The sides of triangle are $a = 32, b = 40, c = 38$ $a=32 , b=40 ,c =38$
Let the angles opposite to sides $a, b, c$ $a,b,c$ are $∠ A, ∠ B, ∠ C$ $/_A ,/_B,/_C$ respetively.
$cosA= (b^2+c^2-a^2)/(2bc)=(40^2+38^2-32^2)/(2*40*38)=0.6644 :./_A= cos^-1 (0.6644)=48.36^0(2dp)$

$cosB= (c^2+a^2-b^2)/(2ca)=(38^2+32^2-40^2)/(2*38*32)=0.3569 :./_B= cos^-1 (0.3569)=69.09^0(2dp)$

$cosC= (a^2+b^2-c^2)/(2ab)=(32^2+40^2-38^2)/(2*32*40)=0.4609 :./_C= cos^-1 (0.4609)=62.55^0(2dp)$

The angles are $/_A=48.36^0, ./_B=69.09^0,/_C= 62.55^0$ [Ans]

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How do you find the measure of each of the angles of a triangle given the measurements of the sides are 32, 40, 38?

1 Answer

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