Question

if an triangle ABC right angled at AB =6 units and BC = 8 units then the find the value of Sin A Cos C + Cost. sinC​

Answer 1

$\huge \underbrace \mathfrak \red{ \star \: {Answer} \: \star}$

Given In ∆ABC, <ABC = 90°

AB = 6 units , BC = 8 units

By Phythogarian theorem:

i) AC² = AB²+BC²

= 6²+8²

= 36+64

= 100

=> AC = √100

=> AC = 10 units.

ii)sinA = BC/AC = 8/10

cosC = BC/AC = 8/10

cosA = AB/AC = 6/10

sinC = AB/AC = 6/10

Now ,

sinAcosC+cosAsinC

$= \frac{8}{10}\times\frac{8}{10}+\frac{6}{10}\times\frac{6}{10}</p><p>\\ = \frac{64}{100}+\frac{36}{100}</p><p>\\= \frac{64+36}{100}</p><p>\\= \frac{100}{100}$

= 1

Therefore,

sinAcosC+cosAsinC =1

HOPE SO IT WILL HELP.....