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Question:

The figure below shows two right angled triangles ∆OAB and ∆OQP with right angles at vertex A and P, respectively, having the common vertex O, The lengths of some of the sides are indicated in the figure. (Note that the figure is not drawn to scale.) AB and OP are parallel. What is ∠QOB?

Explanation:

Consider the figure given below.

In ∆POQ, OQ = $\sqrt{1^{2} + 2^{2}}$ = √5

In ∆AOB, OB = $\sqrt{1^{2} + 3^{2}}$ = √10

In ∆DQB, QB = $\sqrt{2^{2} + 1^{2}}$ = √5

Now, in ∆QOB, OB² = OQ² + QB²

⇒ ∆QOB is a right isosceles triangle, right angles at Q.

∴ ∠QOB = 45°

Hence, option (b).

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