Question

please answer this step by step ​

Answer 1

Answer:

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Answer 2

Answer:

4 cos A - 5 sin A = 0

4 cosA = 5 sin A

4/5= sinA/ cos A

4/5 = tan A -----(1)

if we see in a traingle ABC , according to A ,

AC is base and BC is perpendicular and AC is hypotaneous

tan A = perpendicular / base = BC / AC

from equation (1)

4/5 = BC/AC

hence BC = 4 and AC = 5

now we have to AC by using Pythagoras theorem

AB =

$\sqrt{ac^{2} + {bc}^{2} }$

AB =

$\sqrt{ {4}^{2} + {5}^{2} }$

AB =

$\sqrt{16 + 25}$

AB =

$\sqrt{41}$

now sinA = perpendicular /hypotaneous = BC/AB

sinA = 4/√41

and cos A = base / hypotaneous = AC/AB

cosA = 5/√41

according to question

3sinA- 2cos A/ 5sin A + 3 cos A

put the value of sinA and cos A

3(4/√41) -2(5/√41) / 5(4/√41) + 3(5/√41)

taking LCM

12 -10/√41 / 20+15 /√41

12-10/ 20+15

=2/35