Question

Use the information given in the following figure to evaluate:

`(10)/sin x + (6)/sin y  –  6 cot y`.

 

Answer 1

In ΔADC, Using Pythagorean Theorem

AD² + DC² = AC²

DC² = 20² – 12²

DC² = 256

DC = `sqrt(256)`

∴ DC = 16

Now,

BC = BD + DC

21 = BD + 16

∴ BD = 5

In ΔADB, using Pythagorean Theorem

AD² + BD² = AB²

12² + 5² = AB²

AB² = 169

AB = `sqrt169`

∴ AB = 13

Now,

sin x = `"BD"/"AB" = (5)/(13)`

sin y = `"AD"/"AC" = (12)/(20) = (3)/(5)`

cot y = `"DC"/"AD" = (16)/(12) = (4)/(3)`

Therefore,

`(10)/sin x + (6)/sin y  –  6 cot y`

`= (10)/(5/13) + (6)/(3/5) – 6 (4/3)`

`= 10 xx 13/5 + 6 xx 5/3 - 6(4/3)`

= `(130)/(5) + (30)/(3)  –  (24)/(3)`

= 26 + 10 – 8

= 28