Proof AB=3.6sin0.2 i've done. $\displaystyle \frac{a}{sin0.4} = \frac{1.8}{sin1.37}$ $\displaystyle \rightarrow a=\frac{1.8sin0.4}{sin1.37}$ but duno how to derive the answer. thanks!
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Originally Posted by BabyMilo Proof AB=3.6sin0.2 i've done. $\displaystyle \frac{a}{sin0.4} = \frac{1.8}{sin1.37}$ $\displaystyle \rightarrow a=\frac{1.8sin0.4}{sin1.37}$ but duno how to derive the answer. thanks! Use the cosine rule instead.
Originally Posted by Prove It Use the cosine rule instead. Still I dont understand how i would dervie the required. $\displaystyle a=\sqrt{2*1.8^2-2*1.8*1.8cos0.4}$ They are not whole numbers so how would i find the required. and since this is in cos. thanks!
Is it just, $\displaystyle a=\sqrt{2*1.8^2-2*1.8*1.8cos0.4}$ $\displaystyle a=\sqrt{2*3.24-3.6*1.8(0.921)}$ $\displaystyle a=\sqrt{2*3.24-3.6*1.65}$ Or is there something I'm not getting D:
Originally Posted by BabyMilo Proof AB=3.6sin0.2 i've done. $\displaystyle \frac{a}{sin0.4} = \frac{1.8}{sin1.37}$ $\displaystyle \rightarrow a=\frac{1.8sin0.4}{sin1.37}$ but duno how to derive the answer. thanks! Solve it nevermind. cut the triangle into half. so it's 1.8sin0.2 x 2 =3.6sin0.2 as required. haha.
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