In triangle ABC, right angled at B, if tan A = 6/8, then find the value of sin A. cos C + cos A. sin C
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Given: ABC is a triangle and right angled at B. If tanA=6/8
To prove: Find the value of sinA. cosC+ cosA. sinC.
Proof: ABC is a triangle and right angled at B.
tanA=6/8=P/B
sinA × cosC + cosA × sinC
P/H × B/H + B/H × P/H
sinA=P/H= 6/? = 6/10
By Pythagoras theorem
AB^2 + BC^2 = AC^2
(8) ^2 + (6) ^2 = AC^2
64 + 36 = AC^2
100 = AC^2
10 = AC
The value of sinA × cosC + cosA × sinC
sinA × cosC + cosA × sinC
P/H × B/H + B/H × P/H
6/10 × 6/10 + 6/10 × 6/10
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