The solution of \triangle A B C given that B=45^{\circ}, C=105^{\circ} and c=\sqrt{2} is

Correct Answer is : A=30∘,a=3−1,b=2(3−1)

A=30^{\circ}, a=\sqrt{3}-1, b=\sqrt{2}(\sqrt{3}-1)

A=30^{\circ}, a=\sqrt{3}+1, b=\sqrt{2}(\sqrt{3}-1)

A=30^{\circ}, a=1-\sqrt{3}, b=\sqrt{2}(\sqrt{3}+1)

A=30^{\circ}, a=\sqrt{3}-1, b=\sqrt{2}(\sqrt{3}+1)

NIMCET 2011 Question Paper with solutions

Question : 2 of 120

Marks: +1, -0

The solution of

△ A B C

given that

B=45^{∘}, C=105^{∘}

and

c=√{2}

Solution:

Since,

∠ A+∠ B+∠ C=180^{∘}

∴ \;\; ∠ A=180^{∘}-45^{∘}-105^{∘}=30^{∘}

Using sine rule,

\;{\sin A}/{a}=\;{\sin B}/{b}=\;{\sin C}/{c}

⇒ \;\; \;{\sin 30^{∘}}/{a}=\;{\sin 45^{∘}}/{b}=\;{\sin 105^{∘}}/{√{2}}

⇒ \;{1}/{2 a}=\;{1}/{√{2} b}=\;{\sin 75^{∘}}/{√{2}}=\;{1+√{3}}/{2 √{2} × √{2}}

⇒ \;{1}/{2 a}=\;{1}/{√{2} b}=\;{2}/{4(√{3}-1)}=\;{1}/{2(√{3}-1)}

⇒ \;\; a=√{3}-1, b=√{2}(√{3}-1)

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