An isosceles triangle ABC is right angled at B D is a point inside the triangle ABC P and Q are the ...........

Quantitative Aptitude ( CCS) Geometry

An isosceles triangle ABC is right-angled at B. D is a point inside the triangle ABC. P and Q are the feet of the perpendiculars drawn from D on the side AB and AC respectively of ΔABC. If AP = a cm, AQ = b cm and ∠BAD = 15°, sin 75° = ?

2b /√3a

a/2b

√3a/2b

2a/√3b

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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,
sin 60° = AQ/AD
⇒ √3/2 = b/AD
⇒ AD = 2B/√3
Again from ΔADP,
sin 75° = AP /AD = a/2a√3
= a√3 /2b