Physics, asked by ammijaan, 3 months ago

Plzzz solve the problem.... ​

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Answered by BrainlyEmpire
4

{\Large{\bold{\sf{\underline{Solution}}}}}

  • {\bold{\sf{tanA = \dfrac{3}{4}}}}

  • {\bold{\sf{\dfrac{Side \: opposite \: to \: angle \: X}{Side \: adjacent \: to \: angle \: X} = \dfrac{3}{4}}}}

  • {\bold{\sf{\dfrac{YZ}{XY} = \dfrac{3}{4}}}}

{\bold{\sf{Let,}}}:-

  • {\bold{\sf{YZ \: = \: 3a}}} and {\bold{\sf{XY \: = \: 4a}}}

  • {\bold{\sf{So, \: now \: we \: have \: to \: find \: XZ \: by \: using \: phythagoras \: theorm}}}

(AC)² = (BC)² + (AB)²

(AC)² = (3x)² + (4x)²

(AC)² = (9x) + (16x)

(AC)² = 25x

AC = √25x

AC = 5x

{\small{\bold{\sf{Now}}}}:-

  • sinA = {\bold{\sf{\dfrac{Side \: opposite \: to \: angle \: X}{Hypotenuse}}}}

  • {\bold{\sf{\dfrac{YZ}{XZ}}}}

  • {\bold{\sf{\dfrac{3x}{5x}}}}

  • {\bold{\sf{\dfrac{3}{5}}}}

{\small{\bold{\sf{Now}}}}:-

  • cosA = {\bold{\sf{\dfrac{Side \: adjacent \: to \: angle \: X}{Hypotenuse}}}}

  • {\bold{\sf{\dfrac{XY}{XZ}}}}

  • {\bold{\sf{\dfrac{4x}{5x}}}}

  • {\bold{\sf{\dfrac{4}{5}}}}

  • {\small{\bold{\sf{Now}}}}

  • {\bold{\sf{\dfrac{1}{sinA}}}} + {\bold{\sf{\dfrac{1}{cosA}}}} =

  • {\bold{\sf{\dfrac{\dfrac{1}{3}}{5}}}} + {\bold{\sf{\dfrac{\dfrac{1}{4}}{5}}}} =

  • {\bold{\sf{\dfrac{5}{3}}}} + {\bold{\sf{\dfrac{5}{4}}}} =

  • {\bold{\sf{\dfrac{5 \times 4 + 5 \times 3}{3 \times 4}}}}

  • {\bold{\sf{\dfrac{20 + 15}{12}}}}

  • {\bold{\sf{\dfrac{35}{12}}}}

More Knowledge :-

Trigonometric Identities :-

\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ $\: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1$\end{minipage}}

Trigonometric Table :-

\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & $\dfrac{1}{2} &$\dfrac{1}{\sqrt{2}} & $\dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & $\dfrac{\sqrt{3}}{2}&$\dfrac{1}{\sqrt{2}}&$\dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0&$\dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}

Request:-

  • kindly see the answer from brainly.in website
Answered by ItzMayu
21

Answer:

{\Large{\bold{\sf{\underline{Solution}}}}}

{\bold{\sf{tanA = \dfrac{3}{4}}}}

{\bold{\sf{\dfrac{Side \: opposite \: to \: angle \: X}{Side \: adjacent \: to \: angle \: X} = \dfrac{3}{4}}}}

{\bold{\sf{\dfrac{YZ}{XY} = \dfrac{3}{4}}}}

{\bold{\sf{Let,}}}:-

{\bold{\sf{YZ \: = \: 3a}}} and {\bold{\sf{XY \: = \: 4a}}}

{\bold{\sf{So, \: now \: we \: have \: to \: find \: XZ \: by \: using \: phythagoras \: theorm}}}

(AC)² = (BC)² + (AB)²

(AC)² = (3x)² + (4x)²

(AC)² = (9x) + (16x)

(AC)² = 25x

AC = √25x

AC = 5x

{\small{\bold{\sf{Now}}}}:-

sinA = {\bold{\sf{\dfrac{Side \: opposite \: to \: angle \: X}{Hypotenuse}}}}

{\bold{\sf{\dfrac{YZ}{XZ}}}}

{\bold{\sf{\dfrac{3x}{5x}}}}

{\bold{\sf{\dfrac{3}{5}}}}

{\small{\bold{\sf{Now}}}}:-

cosA = {\bold{\sf{\dfrac{Side \: adjacent \: to \: angle \: X}{Hypotenuse}}}}

{\bold{\sf{\dfrac{XY}{XZ}}}}

{\bold{\sf{\dfrac{4x}{5x}}}}

{\bold{\sf{\dfrac{4}{5}}}}

{\small{\bold{\sf{Now}}}}

{\bold{\sf{\dfrac{1}{sinA}}}} + {\bold{\sf{\dfrac{1}{cosA}}}} =

{\bold{\sf{\dfrac{\dfrac{1}{3}}{5}}}} + {\bold{\sf{\dfrac{\dfrac{1}{4}}{5}}}} =

{\bold{\sf{\dfrac{5}{3}}}} + {\bold{\sf{\dfrac{5}{4}}}} =

{\bold{\sf{\dfrac{5 \times 4 + 5 \times 3}{3 \times 4}}}}

{\bold{\sf{\dfrac{20 + 15}{12}}}}

{\bold{\sf{\dfrac{35}{12}}}}

More Knowledge :-

Trigonometric Identities :-

\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ $\: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1$\end{minipage}}

Trigonometric Table :-

\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & $\dfrac{1}{2} &$\dfrac{1}{\sqrt{2}} & $\dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & $\dfrac{\sqrt{3}}{2}&$\dfrac{1}{\sqrt{2}}&$\dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0&$\dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}

Request:-

kindly see the answer from brainly.in website

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