Question

IN A TRIANGLE ABC RIGHT ANGLED AT B , AB =6 UNITS AND BC=8 THEN FIND THE VALUE OF sinA.cosC+cosA.sinC

Answer 1

I think this is the answer

Answer 2

Solution:

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

= $\frac{8}{10}\times\frac{8}{10}+\frac{6}{10}\times\frac{6}{10}$

= $\frac{64}{100}+\frac{36}{100}$

= $\frac{64+36}{100}$

= $\frac{100}{100}$

= $1$

Therefore,

sinAcosC+cosAsinC =1

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