Question

In the given figure;

BC = 15 cm and sin B = `(4)/(5)`

1. Calculate the measure of AB and AC.
2. Now, if tan ∠ADC = 1; calculate the measures of CD and AD.

Also, show that: tan²B - `1/cos^2 "B" = – 1 .`

Answer 1

Given 

sin B = `(4)/(5)`

Now, sin B = `"P"/"H"`

`4/5 = "AC"/"AB"`

Therefore AC = 4x, and AB = 5x

Since

BC² + AC² = AB²  ...[ Using Pythagoras Theorem]

(5x)² – (4x)² = BC

BC² = 9x²

∴ BC = 3x

Now

BC = 15

3x = 15

x = 5

(i) AC = 4x
  = 4 x 5
  = 20 cm

And
 AB = 5x 
= 5 x 5
= 25 cm

(ii) Given 
tan ∠ADC = `(1)/(1)`

i.e. `"perpendicular"/"base" = "AC"/"CD" = (1)/(1)`

Therefore if length of perpendicular = x, length of hypotenuse = x

Since
AC² + CD² = AD²  ...[Using Pythagoras Theorem]

(x)2 – (x)² = BC

AD² = 2x²

∴ AD = `sqrt2x`

Now 

AC = 20
x = 20

So
AD = `sqrt2x`
= `sqrt2` x 20

= 20`sqrt2"cm"`

And
CD = 20 cm

Now 
tan B = `"AC"/"BC" = (20)/(15) = (4)/(3)`

cos B = `"BC"/"AB" = (15)/(25) = (3)/(5)`

So
tan² B – `1/cos^2 B`

=`(4/3)^2 –  1/(3/5)^2`

= `(16)/(9) –  (25)/(9)`

= `– (9)/(9)`

= –  1