How do you solve the triangle given $A = 55^{\circ}, b = 3, c = 10$ $A=55^circ, b=3, c=10$ ?

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Answer 1 · 2018-02-01T14:59:18

$a = 8.64$ $a=8.64$ units long

I'm assuming that you are trying to find side $a$ $a$ of the triangle.

Well, the cosine rule states that

$a^{2} = b^{2} + c^{2} - 2 b c cos A$ $a^2=b^2+c^2-2bc cos A$

Plugging in $b = 3$ $b=3$ , $c = 10$ $c=10$ , $A = 55^{\circ}$ $A=55^@$ , we get

$a^{2} = 9 + 100 - 60 {cos 55}^{\circ}$ $a^2=9+100-60 cos55^@$

$a^{2} = 109 - 34.41$ $a^2=109-34.41$

$a^{2} = 74.59$ $a^2=74.59$

$a = \sqrt{74.59} \approx 8.64$ $a=sqrt(74.59)~~8.64$

So, side $a$ $a$ will be $8.64$ $8.64$ units long.

How do you solve the triangle given A=55∘,b=3,c=10A=55^circ, b=3, c=10?