**Answer:** Option **
C **

**Explanation:**

Given, ΔABC is an isosceles trialge and ∠B is right-angled.

∴ ∠A = ∠C

and ∠B = 90°

We know that the sum of the angles of a triangle is 180°.

∴ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + ∠A = 180° [∵ ∠B = 90° & ∠A = ∠C]

⇒ 2∠A = 180° – 90° = 90°

⇒ ∠A = 45° and ∠C = 45° [∵ ∠A = ∠C]

From ΔADP,

∠APD + ∠PAD + ∠ADP = 180° |
[ |
∠BAD = 15° (given) |
] |

∴ ∠PAD = 15° |

⇒ 90° + 15° + ∠ADP = 180° [PD ⊥ AB ∴ ∠APD = 90°]

⇒ ∠ADP = 180° – 90° – 15° = 75°

Now, ∠A = ∠BAD + ∠DAC

⇒ 45° = 15° + ∠DAC

⇒ ∠DAC = 45° – 15° = 30°

∴ ∠DAQ = 30°

From ΔADQ,

∠AQD + ∠DAQ + ∠ADQ = 180°

⇒ 90° + 30° + ∠ADQ = 180° [DQ ⊥ AC ∴ ∠AQD = 90°]

⇒ ∠ADQ = 180° – 90° – 30° = 60°

Again from ΔADQ,

Again from ΔADP,