Question

5. find find 'x' and 'y' in each case​

Answer 1

in right ∆ ABC

tan 60° = AB/BC

√3 = y/x

y = √3x

in right angle ABD

tan 30° = AB/BD

1/√3 = y/x+20

x + 20 = y √3

now i am going to put the value of y

√3x + 20 = √3x(√3)

√3 + 20 = 3x

x = √3(1 + 20)/3

x = √3(21)/3

x =√3(7)

x = 7√3

y = √3 x

=√3 [(7√3)]

= 7(3)

=21

2 no.

in right ∆ BCD

tan 45° = BC/CD

1 = x/y

y = x

in right ∆ ABD

tan 60° = AB/CD

√3 = 20 + x /y

y√3 = 20 +x

now i am going to put the value of y

x√3 = 20 + x

x√3 - x = 20

X(√3 - 1) =20

x = 20/(√3 - 1)

x = 20/(√3 - 1)×(√3+ 1)/(√3+1)

x = 20(√3+1)/(√3)^2 -1^2

x = 20(√3+1)/3-1

x = 20(√3+1)/2

x = 10(√3+1)

x = y

y = 10(√3+1)