if tan tita is 1/2 then evaluate 2 sin theta + 3cos theta upon 4 costheta + 3 sintheta
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given :-
tan∅ = 1/2
let us consider any right angle triangle ABC right angle at C
we know that tan∅ = perpendicular/base
= BC/AC
therefore BC = 1cm and AC = 2cm
by Pythagoras theorem we get,
➡ AB² = BC² + AC²
➡ AB² = 1² + 2²
➡ AB² = 5
➡ AB = √5
hence, the base is √3cm
now, sin∅ = perpendicular/hypotenuse
= BC/AB
= 1/√5
cos∅ = base/hypotenuse
= AC/AB
= 2/√5
hence value of (2sin∅ + 3cos∅)/(4cos∅ + 3sin∅)
= [2(1/√5) + 3(2/√5)]/[4(2/√5) + 3(1/√5)]
= (2/√5 + 6/√5)/(8/√5 + 3/√5)
= (8/√5)/(11/√5)
= 8/√5 × √5/11
= 8/11 FINAL ANSWER
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