Geometry Practice Question for SSC CGL, SSC CHSL, Banking

Question of The Day27-07-2020

In the given figure, what will be the measure of the ∠ECB – ∠ BAC? (Given: O is the centre of the circle)

Answer

Correct Answer : a ) 15^o

Explanation :

In the given figure,

We know that the angle subtended on the circumference of the circle is half the angle subtended at the centre of the circle by the same arc.

So,

∠BAC=\({1 \over 2}\)∠BOC=\({1 \over 2}\)*70=35^o………1

Similarly,

⇒∠DCA=\({1 \over 2}\)∠DOA=\({1 \over 2}\)*80=40^o

We know that, angle subtended by the semicircle/diameter on the circumference is 90 degree

So, ∠BCA = 90^o

Now, ∠ ECB = ∠ BCA – ∠ DCA = 90^o – 40^o = 50^o ………. (2)

Therefore, from equation 1 and 2

∠ECB – ∠ BAC = 50^o – 35^o = 15^o

Hence, (a) is the correct answer.

Such type of question is asked in various government exams like SSC CGL, SSC MTS, SSC CPO, SSC CHSL, RRB JE, RRB NTPC, RRB GROUP D, RRB OFFICER SCALE-I, IBPS PO, IBPS SO, RRB Office Assistant, IBPS Clerk, RBI Assistant, IBPS RRB OFFICER SCALE 2&3, UPSC CDS, UPSC NDA etc.

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