Question

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).​

Answer 1

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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