Solve for angleTβ³π ππ, given π=1.2ππ, π =2.4ππ, π‘=2.2ππ
Ans:-
According to the question we solve for angle β t\angle tβ t
So, cosβ tcos\angle tcosβ t === r2+s2βT22sr\dfrac{r^2+ s^2 -T^2}{2sr}2srr2+s2βT2β
β1.22+2.42β2.222Γ2.4Γ1.2\Rightarrow \dfrac{1.2^2+2.4^2-2.2^2}{2\times2.4\times1.2}β2Γ2.4Γ1.21.22+2.42β2.22β
β\Rightarrowβ cosβ t=cos\angle t =cosβ t= 0.41
β\Rightarrowβ β t\angle tβ t === 1.148
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