BPSC 45th Exam Prelims 2003 Solved Paper

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Question : 121 of 150

Marks: +1, -0

In the following figure ABCD is a square and∠EDC is an equilateral triangle. Points A and E are joined by a straight line. Find ∠DAE.

15°
30°
10°
20°

Solution:

∠ {ADE}=∠ {ADC}+∠ {CDE}

=90^{∘}+60^{∘}

∠ {ADE}=150^{∘}

{DC}= {DE}[ ∵

Sides of Equilateral

∆

]

{DC}= {AD}[ ∵

Sides of Square

]

∴ A D=D E

⇒ ∠ {DAE}=∠ {DEA}

Now, In

∆ {ADE}

∠ {ADE}+∠ {DAE}+∠ {DEA}=180^{∘}

150^{∘}+2 ∠ {DAE}=180^{∘}

2 ∠ {DAE}=30^{∘}

∠ {DAE}=15^{∘}

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