In a triangle ABC, AB= 20, AC = 21 and BC= 29. The points D and Elie on the line segment BC, with BD = 8 and EC = 9. Find angleDAE (in degree measure).
Answers
Given : In a triangle ABC, AB= 20, AC = 21 and BC= 29.
The points D and E lie on the line segment BC, with BD = 8 and EC = 9.
To Find : angle DAE (in degree measure).
Solution:
AB= 20, AC = 21 and BC= 29
=> 20² + 21² = 29²
=> AB² + AC² = BC²
Using Converse of Pythagoras theorem
BAC is right angles triangle at A
Cos B = 20/29
Cos C = 21/29
in Δ ABD
AD² = AB² + BD² - 2 (AB(BD) CosB
=> AD² = 20² + 8² - 2(20)(8) (20/29)
=> AD² = 464 - 6400/29
=. AD² = 7056/29
=> AD = 84/√29
AE² = AC² + CE² - 2(AC)( CE) CosC
=>AE² = 21² + 9² - 2(21)(9) (21/29)
=> AE² = 522 - 7,938/29
=> AE² = 7200/29
=> AE = 60 √2 / √29
BC = BD + DE + EC
=> 29 = 8 + DE + 9
=> DE = 12 cm
DE² = AD² + AE² - 2(AD) (DE) Cos ∠DAE
=> 12² = 7056/29 + 7200/29 - 2 ( 84/√29)(60 √2 / √29) Cos ∠DAE
=> 144 = 14256/29 - (10,080√2 / 29)Cos ∠DAE
=> 4,176 = 14256 - 10,080√2 Cos ∠DAE
=> - 10,080= - 10,080√2 Cos ∠DAE
=> 1 = √2 Cos ∠DAE
=> Cos ∠DAE = 1/√2
=> ∠DAE = 45°
Hence angle DAE (in degree measure). = 45°
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