Math Analysis: Solve the triangle. C=37degrees; a=5.2; b=6
use cosine rule
\(a^2=b^2+c^2-2bc \cos A\)
that formula doesn't look familiar to me
i know ofc a^2=b^2 + c^2 but not the "-2bc cosA"
and this is not a right triangle
r u searching area of triangle
That's why we use cosine rule
no, just solving the triangle
that's the cosine rule?
ya..... got 3 actually but they all the same for this question \(c^2=a^2+b^2-2ab\cos C\)
oh, it does sound familiar..
just substitute the values and you will get the length of c then you can find all others angle and length.
well if it is not familiar to u then at first just find out the area of triangle by using the very simple formula ie A=1/2 ab sinC then find out the value of c by with another formula ie A= root under s(s-a)(s-b)(s-c) where s is a+b+c/2
ok, im having a little trouble with finding cosine A
but cosine law is also one of the formula and may be u'll be familiar with it later
well there r other formula of cosine laws
you can also use sine rule \(\Huge\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\)
like foe this one u use 2abcosC= a^2+b^2-c^2
or \(\Huge\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)
but in order to use that i need to have angle and a side of either A, B, or C right?
you are given a, b and C so simply use 2abcosC= a^2+b^2-c^2 to find c
well i got c=3.63
ok. how about the degrees of A im getting the wrong answer :\
i got A=60
it's 59.55 but how did u get that answer
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