2--3और 4--5 केमध्य तीनतीन परिमेय संख्या ज्ञात करें
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Answers
Answer:
We have given that,
\displaystyle{\sf\:3\:\cot\:A\:=\:4}3cotA=4
We have to find the value of
\displaystyle{\sf\:\dfrac{3\:\cos\:A\:+\:2\:\sin\:A}{3\:\cos\:A\:-\:2\:\sin\:A}}
3cosA−2sinA
3cosA+2sinA
Now,
\displaystyle{\sf\:3\:\cot\:A\:=\:4}3cotA=4
\displaystyle{\implies\boxed{\sf\:\cot\:A\:=\:\dfrac{4}{3}}}⟹
cotA=
3
4
Consider a right triangle ABC right angled at angle B.
We know that,
\displaystyle{\pink{\sf\:\cot\:A\:=\:\dfrac{Adjacent\:side}{Opposite\:side}}}cotA=
Oppositeside
Adjacentside
\displaystyle{\implies\sf\:\dfrac{4}{3}\:=\:\dfrac{AB}{BC}}⟹
3
4
=
BC
AB
\displaystyle{\implies\sf\:AB\:=\:4x}⟹AB=4x
\displaystyle{\sf\:BC\:=\:3x}BC=3x
Now, by Pythagoras theorem,
\displaystyle{\pink{\sf\:(\:AC\:)^2\:=\:(\:AB\:)^2\:+\:(\:BC\:)^2}}(AC)
2
=(AB)
2
+(BC)
2
\displaystyle{\implies\sf\:AC^2\:=\:(\:4x\:)^2\:+\:(\:3x\:)^2}⟹AC
2
=(4x)
2
+(3x)
2
\displaystyle{\implies\sf\:AC^2\:=\:16x^2\:+\:9x^2}⟹AC
2
=16x
2
+9x
2
\displaystyle{\implies\sf\:AC^2\:=\:25x^2}⟹AC
2
=25x
2
\displaystyle{\implies\sf\:AC\:=\:5x\:}⟹AC=5x
\displaystyle{\therefore\boxed{\sf\:Hypotenuse\:=\:5x}}∴
Hypotenuse=5x
Now, we know that,
\displaystyle{\pink{\sf\:\sin\:A\:=\:\dfrac{Opposite\:side}{Hypotenuse}}}sinA=
Hypotenuse
Oppositeside
\displaystyle{\implies\sf\:\sin\:A\:=\:\dfrac{3\:\cancel{x}}{5\:\cancel{x}}}⟹sinA=
5
x
3
x
\displaystyle{\implies\sf\:\sin\:A\:=\:\dfrac{3}{5}}⟹sinA=
5
3
\displaystyle{\implies\sf\:2\:\sin\:A\:=\:\dfrac{2\:\times\:3}{5}}⟹2sinA=
5
2×3
\displaystyle{\implies\boxed{\blue{\sf\:2\:\sin\:A\:=\:\dfrac{6}{5}}}}⟹
2sinA=
5
6
Now, we know that,
\displaystyle{\pink{\sf\:\cos\:A\:=\:\dfrac{Adjacent\:side}{Hypotenuse}}}cosA=
Hypotenuse
Adjacentside
\displaystyle{\implies\sf\:\cos\:A\:=\:\dfrac{4\:\cancel{x}}{5\:\cancel{x}}}⟹cosA=
5
x
4
x
\displaystyle{\implies\sf\:\cos\:A\:=\:\dfrac{4}{5}}⟹cosA=
5
4
\displaystyle{\implies\sf\:3\:\cos\:A\:=\:\dfrac{3\:\times\:4}{5}}⟹3cosA=
5
3×4
\displaystyle{\implies\boxed{\green{\sf\:3\:\cos\:A\:=\:\dfrac{12}{5}}}}⟹
3cosA=
5
12
Now, we have to find the value of
\displaystyle{\sf\:\dfrac{3\:\cos\:A\:+\:2\:\sin\:A}{3\:\cos\:A\:-\:2\:\sin\:A}}
3cosA−2sinA
3cosA+2sinA
\displaystyle{\implies\sf\:\dfrac{\dfrac{12}{5}\:+\:\dfrac{6}{5}}{\dfrac{12}{5}\:-\:\dfrac{6}{5}}}⟹
5
12
−
5
6
5
12
+
5
6
\displaystyle{\implies\sf\:\dfrac{\dfrac{12\:+\:6}{5}}{\dfrac{12\:-\:6}{5}}}⟹
5
12−6
5
12+6
\displaystyle{\implies\sf\:\dfrac{\dfrac{18}{5}}{\dfrac{6}{5}}}⟹
5
6
5
18
\displaystyle{\implies\sf\:\dfrac{18}{\cancel{5}}\:\times\:\dfrac{\cancel{5}}{6}}⟹
5
18
×
6
5
\displaystyle{\implies\sf\:\cancel{\dfrac{18}{6}}}⟹
6
18
\displaystyle{\implies\sf\:3}⟹3
\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\dfrac{3\:\cos\:A\:+\:2\:\sin\:A}{3\:\cos\:A\:-\:2\:\sin\:A}\:=\:3\:}}}}∴
3cosA−2sinA
3cosA+2sinA
=3
hello you answer is here dear
14
hope it is helpful
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