Question

In the adjoining figure, 

i. Mn (arc CE) = 54° , m (arc BD) = 23° , find measure of ∠CAE. 

ii. If AB = 4.2,BC = 5.4, AE = 12.0, find AD. 

iii. If AB = 3.6, AC = 9.0, AD = 5.4, find AE.

Answer 1

i. Chords BC and ED intersect each other externally at point A. 

∴ ∠CAE = 1/2 [m(arc CE) – m(arc BD)] 

= 1/2 (54° – 23°) 

= 1/2 × 31° 

∴ m∠CAE = 15.5° 

ii. AC = AB + BC [A – B – C] 

= 4.2 + 5.4 

= 9.6 units 

Now, AB × AC = AD × AE [Theorem of external division of chords] 

∴ 4.2 × 9.6 = AD × 12.0

∴ AD =  (4.2 x 9.6) / 12

∴ AD = 3.36 units 

iii. AB × AC = AD × AE [Theorem of external division of chords] 

∴ 3.6 × 9.0 = 5.4 × AE 

∴ AE = (3.6 x 9.0) / 5.4

∴ AE = 6 units