Question

If $\alpha , \beta, \gamma$ are positive acute angles, such that:
$sin( \alpha + \beta - \gamma ) = \dfrac{1}{2}$,
$cos( \beta + \gamma - \alpha ) = \dfrac{1}{2}$,
$tan( \gamma + \alpha - \beta ) = 1$,
Then, find the values of $\alpha , \beta, \gamma$.
⇒ Concept: Introduction to Trigonometry
⇒ Class: 10
⇒ Thanks for helping me and kindly don't spam!

Answer 1

Answer:

${ \underline{ \large{ \pmb{ \sf{Given... </p><p>}}}}}$

• ${ \sf{sin( \alpha + \beta \: - \gamma) = \frac{1}{2} }}$
• ${ \sf{cos( \beta + \gamma - \alpha) = \frac{1}{2} }}$
• ${ \sf{tan( \gamma + \alpha - \beta) = 1}}$

${ \underline{ \large{ \pmb{ \sf{Find... }}}}}$

• ${ \sf{Values \: of \: \alpha, \: \beta, \: \: \gamma}}$

${ \underline{ \large{ \pmb{ \sf{Solution...}}}}}$

From Question,

: ➙ Sin(α + β - γ) = 1/2

: ➙ Sin (α + β - γ) = Sin30°

: ➙ α + β - γ = 30° .......(1)

____________________

: ➙ Cos(β + γ - α) = 1/2

: ➙ Cos(β + γ - α) = cos60°

: ➙ β + γ - α = 60° ......(2)

___________________

: ➙ Tan(γ + α - β) = 1

: ➙ Tan(γ + α - β) = Tan45°

: ➙ γ + α - β = 45° .......(3)

_________________

Now By adding equation (1) & (2):

${ \implies{ \sf { α + β - γ + β + γ - α = 30 + 60}}} \\ \\ { \implies{ \sf{2β = 90}}} \\ \\ { \implies{ \sf{β = 45° }}}$

Now By Adding Equation (2) & (3) :

${ \implies{ \sf{β + γ - α - β + γ + α = 60 + 45 }}} \\ \\ { \implies{ \sf{2 γ = 105}}} \\ \\ { \implies{ \sf{ γ = 55 \frac{1}{2} }}}$

Now substitute both values in equation (1) :

${ \implies{ \sf{α + β - γ = 30°}}} \\ \\ { \implies{ \sf{α + 45 - 55 \frac{1}{2} = 30}}} \\ \\ { \implies{ \sf{α = \frac{60 + 105 - 90}{2} }}} \\ \\ { \implies{ \sf{α = 37 \frac{1}{2} }}}$

Therefore,

• α = 37½°
• β = 46°
• γ = 55½°

Answer 2

$\large \underbrace{\pmb{\bf{\orange{Given:}}}}$

Given that, $\sf{\alpha, \beta, \gamma}$ are positive acute angles such that :

⇒$\sf sin(\alpha+\beta-\gamma)=\dfrac{1}{2}$

⇒$\sf cos(\beta+\gamma-\alpha)=\dfrac{1}{2}$

⇒$\sf tan(\gamma+\alpha-\beta)=1$

$\large\underbrace{\pmb{\bf{\orange{To~find:}}}}$

We have to find the values of $\sf {\alpha,\beta,\gamma}$.

$\large \underbrace{\pmb{\bf{\orange{Solution:}}}}$

Here,

⇒$\sf sin(\alpha+\beta-\gamma)=\dfrac{1}{2}:sin30\degree ....[i]$

⇒$\sf cos(\beta+\gamma-\alpha=\dfrac{1}{2}:cos60\degree ....[ii]$

⇒$\sf tan(\gamma+\alpha-\beta)=1:tan45\degree ....[iii]$

Now, adding [ii] and [iii] for knowing the value of $\sf\gamma$ :

⇒$\sf (\beta+\gamma-\alpha)+(\gamma+\alpha-\beta)=(60\degree)+(45\degree)$

⇒$\sf \beta-\beta+\gamma+\gamma-\alpha+\alpha=105\degree$

⇒$\sf 2\gamma=105\degree$

${\pmb{\sf{\purple{Therefore,~\gamma=52.5\degree}}}}$

Now, adding [i] and [ii] for knowing the value of $\beta$ :

⇒$\sf (\alpha+\beta-\gamma)+(\beta+\gamma-\alpha)=(30\degree)+(60\degree)$

⇒$\sf \alpha-\alpha+\beta+\beta-\gamma+\gamma=90\degree$

⇒$\sf 2\beta=90\degree$

${\pmb{\sf{\purple{Therefore,~\beta=45\degree}}}}$

In [iii], substituting the values of $\sf \gamma,\beta$ for knowing the value of $\sf \alpha$ :

⇒$\sf \gamma+\alpha-\beta=45\degree$

⇒$\sf 52.5\degree + \alpha -45\degree=45\degree$

⇒$\sf 7.5+\alpha=45\degree$

${\pmb{\sf{\purple{Therefore,~\alpha=37.5\degree}}}}$

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