Question

1) if 5 cot ø =4 so ,, ( 5 sin ø - 3 cos Ø )

Answer 1

From the interior point O of an equilateral ABC, the sides bc , ac , and

From the interior point O of an equilateral ABC, the sides bc , ac , andPerpendiculars OD, OE and OF are drawn on ab, respectively, the lengths of these perpendiculars are 24 cm, 20, 22 cm respectively, find the area of

Step-by-step explanation:

Let each side of ㎝ equilateral triangle ABC be

′

a

′

㎝

Now, ar△OAB=

2

1

×AB×OP=

2

1

×a×14=7a㎠→1

ar△OBC=

×BC×OQ

=

2

1

×a×10=5a㎠→2

ar△OAC=

2

1

×AC×OR=

2

1

×a×6=3a㎠→3

∴ar△ABC=1+2+3=7a+5a+3a=15a㎠

Also area of equilateral triangle ABC=

4

3

a

2

Now,

4

3

a

2

=15a⇒a=

3

15×4

×

3

3

=

3

60

3

=20

3

㎝

Now, ar△ABC=

4

3

×(20

3

)

2

=300

3 ㎠

Answer 2

1 -) ( 5 sin ø - 3 cos Ø )

________________

sin Ø + 2 cos Ø

← SOLUTION→

( 5 sin ø - 3 cos Ø )

________________

sin Ø + 2 cos Ø

=> 5 - 3 cot ∅

( ___________ )

1 + 2 cot ∅

=>. 5 - 3 × 4

[ ____. ]

5

________________

1 + 2 x 4

__

5

= 25 - 12

________. (:. cot Ø = 4 / 5

5 + 8

= 13

___

13

= 1

2 - ) ( b sin Ø - a cos Ø)

_______________

b sin Ø+ a cos Ø

SOLUTION

( b sin Ø - a cos Ø)

_______________

b sin Ø+ a cos Ø

= ( b tan Ø - a )

____________

( b tan Ø+ a )

b

= b × ___ - a

a

__________. (:. tan Ø= b /a

b

b × ___. + a

a

= b ² - a ²

________

b² + a ²