Question

$1) \: in \: a \: triangle \: abc \: if \: a = 2 \sqrt{2} \: b = 2 \sqrt{3 \: }c = 75degree . \find\: the \:order \: other \: sides \: and \: the \: angle$



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Answer 1

Question :-

In ∆ ABC,

$\bf \:if \: a = 2 \sqrt{2} \:and \: b = 2 \sqrt{3 \: }$

 ∠C = 75°. Find the other sudes and angle of a triangle.

$\huge \red{AηsωeR } ✍$

$\large \red{\bf \:  Given :- } ✍$

A ∆ ABC, in which ∠C = 75° and two sides are

$\bf \:a = 2 \sqrt{2} \:and \: b = 2 \sqrt{3 \: }$

$\large \red{\bf \:  To  \: Find :- } ✍$

• Side c of triangle ABC
• ∠A and ∠B of triangle ABC.

Formula Used :-

Cosine Law :-

$\bf \:cosC = \dfrac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab}$

Sine Law :

$\bf \:\dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC} = k$

$\large \red{\bf \:  Solution :- } ✍$

$\bf \:  ⟼ cos75° = cos(45° + 30°)$

$\bf \:  ⟼ cos75° = cos45°cos30° - sin45°sin30°$

$\bf \:  ⟼ cos75° = \dfrac{ \sqrt{3} }{2} \times \dfrac{1}{ \sqrt{2} } - \dfrac{1}{ \sqrt{2} } \times \dfrac{1}{2}$

$\bf \:  ⟼ cos75° = \dfrac{ \sqrt{3} - 1}{2 \sqrt{2} } \: ⟼ \: (1)$

$\bf \:  ⟼ sin75° = sin(45° + 30°)$

$\bf \:  ⟼ sin75° = sin45°cos30° + cos45°sin30°$

$\bf \:  ⟼ sin75° = \dfrac{ 1}{2} \times \dfrac{1}{ \sqrt{2} } + \dfrac{ \sqrt{3} }{2} \times \dfrac{1}{ \sqrt{2} }$

$\bf \:  ⟼ sin75° = \dfrac{ \sqrt{3} + 1}{2 \sqrt{2} } \: ⟼ \: (2)$

$\begin{gathered}\bf\red{According\:to\:the\:question,} \end{gathered}$

$\bf \:  ⟼ a = 2 \sqrt{2} \:and \: b = 2 \sqrt{3 \: } \: and \: ∠C = 75°$

Using Cosine Law, we get

$\bf \:cosC = \dfrac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab}$

On substituting the values of a, b and ∠C, we get

$\bf ⟼ \:cos75°= \dfrac{ {(2 \sqrt{2}) }^{2} + {(2 \sqrt{3} )}^{2} - {c}^{2} }{2 \times 2 \sqrt{2} \times 2 \sqrt{3} }$

$\bf ⟼ \:\dfrac{ \sqrt{3} - 1}{2 \sqrt{2} } = \dfrac{ 8 + 12- {c}^{2} }{8 \times \sqrt{2} \times \sqrt{3} }$

$\bf \:  ⟼ ( \sqrt{3} - 1) \times 4 \sqrt{3} = 20 - {c}^{2}$

$\bf \:  ⟼ 12 - 4 \sqrt{3} = 20 - {c}^{2}$

$\bf \:  ⟼ {c}^{2} = 8 + 4 \sqrt{3}$

$\bf \:  ⟼ {c}^{2} = 2(4 + 2 \sqrt{3} )$

$\bf \:  ⟼ {c}^{2} = 2(3 + 1 + 2 \sqrt{3} )$

$\bf \:  ⟼ {c}^{2} = 2 {( \sqrt{3} + 1) }^{2}$

$\bf \:  ⟼ c = \sqrt{2} ( \sqrt{3} + 1) \: ⟼ \: (3)$

$\begin{gathered}\bf\pink{Now \: using \: sine \: law}\end{gathered}$

$\bf \:  ⟼ \bf \:\dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC}$

On substituting the values of a, b, c and ∠C, we get

$\bf⟼ \:\dfrac{2 \sqrt{2} }{sinA} = \dfrac{2 \sqrt{3} }{sinB} = \dfrac{ \sqrt{2} ( \sqrt{3} + 1) }{sin75°}$

$\bf⟼ \:\dfrac{2 \sqrt{2} }{sinA} = \dfrac{2 \sqrt{3} }{sinB} = \dfrac{ \sqrt{2} ( \sqrt{3} + 1) }{\dfrac{ \sqrt{3} + 1}{2 \sqrt{2} } }$

$\bf⟼ \:\dfrac{2 \sqrt{2} }{sinA} = \dfrac{2 \sqrt{3} }{sinB} = 4$

$\bf \:  ⟼ sinA = \dfrac{2 \sqrt{2} }{4} \: and \: sinB = \dfrac{2 \sqrt{3} }{4}$

$\bf \:  ⟼ sinA = \dfrac{1}{ \sqrt{2} } \: and \: sinB = \dfrac{ \sqrt{3} }{2}$

$\large \red{\bf \:  ⟼ A = 45° \: and \: B = 60°} ✍$

$\begin{gathered}\bf\blue{Hence} \end{gathered}$

$\ \red{\bf \: A = 45° \: and \: B = 60° \: and \: c \: = \sqrt{2}( \sqrt{3 + 1)} }$