HEIGHT AND DISTANCE ALL CONCEPT IN ONE GO

Trigonometry:

I.   sin  =

Perpendicular

=

AB

;

Hypotenuse

OB

ii.   cos  =	Base	=	OA	;
	Hypotenuse		OB

iii.  tan  =	Perpendicular	=	AB	;
	Base		OA

iv.  cosec  =	1	=	OB	;
	sin		AB

v.   sec  =	1	=	OB	;
	cos		OA

vi.  cot  =	1	=	OA	;
	tan		AB

Trigonometrical Identities:

i.        sin²  + cos²  = 1.

ii.        1 + tan²  = sec² .

iii.        1 + cot²  = cosec² .

Values of T-ratios:

	0°	(/6) 30°	(/4) 45°	(/3) 60°	(/2) 90°
sin	0		1 √2	√3 2	1
cos	1	√3 2	1 √2		0
tan	0	1 √3	1	√3	not defined

     Angle of Elevation:

Suppose a man from a point O looks up at an object P, placed above the level of his eye. Then, the angle which the line of sight makes with the horizontal through O, is called the angle of elevation of P as seen from O.

 Angle of elevation of P from O = AOP.

Angle of Depression:

Suppose a man from a point O looks down at an object P, placed below the level of his eye, then the angle which the line of sight makes with the horizontal through O, is called the angle of depression of P as seen from O.

Q1-Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:

A. 173 m B. 200 m C. 273 m D. 300 m Explanation: Let AB be the lighthouse and C and D be the positions of the ships. Then, AB = 100 m, ACB = 30° and ADB = 45°. AB = tan 30° = 1          AC = AB x 3 = 100ROOT3 m. AC 3 AB = tan 45° = 1          AD = AB = 100 m. AD  CD = (AC + AD) = (100√3 + 100) m = 100(√3 + 1) = (100 x 2.73) m = 273 m.

Q2-The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

A. 2.3 m B. 4.6 m C. 7.8 m D. 9.2 m Explanation: Let AB be the wall and BC be the ladder. Then, ACB = 60º and AC = 4.6 m. AC = cos 60º = 1 BC 2  BC = 2 x AC = (2 x 4.6) m = 9.2 m.