Solve for angleTβ³π ππ, givenΒ π=1.2ππ,Β π =2.4ππ,Β π‘=2.2ππ
Ans:-
According to the question we solve for angle "\\angle t"
So, "cos\\angle t" "=" "\\dfrac{r^2+ s^2 -T^2}{2sr}"
"\\Rightarrow \\dfrac{1.2^2+2.4^2-2.2^2}{2\\times2.4\\times1.2}"
"\\Rightarrow" "cos\\angle t =" 0.41
"\\Rightarrow" "\\angle t" "=" 1.148
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