QUESTION BASED ON TRIGONOMETRY
Questions based on trigonometry:
In a triangle ABC, right angled at A; if AB= 12cm, AC=5cm. Find all the trigonometric ratios of angle B and C?
2. In a triangle ABC right angle that be. If AC equals 5 centimeters, BC equals 3 centimetre.Find
Sin A
Cos C
Tan A
Cosec C
Sec^2 A - Tan^2 A
3) In a right-angled triangle. If angle A is acute and cot A= 4/3. Find the remaining trigonometric ratio.
Given: CotA = 4/3
Base (AB) = 4
Perpendicular (BC) = 3
By using Pythagoras theorem
AC^2 =AB^2+BC^2
= 42 + 32
= 16 +9
AC^2= 25
AC = 5
Sin A = Perpendicular/Hypotenuse = 3/5
Cos A = Base/Hypotenuse= 4/5
Tan A = Perpendicular/Base=3/4
Cosec A = Hypotenuse/Perpendicular= 5/3
Sec A = Hypotenuse/Base= 5/4
4) For the given figure, if cosˠ=513 and cosβ= 35 Find the length of BD?
In ABC, Cosˠ = 5/13 =BC/AC
If BC= 5k, AC= 13k
AC^2= BC^2+ BC^2
(13k) ^2= 12^2+ (5k) ^2
169 k^2 = 144 + 25 k^2
169k^2-25k^2 = 144
144 k^2= 144
k= 1
BC =5k
BC = 5m
In CDE, Cos β = 3/5 = CD/CE
If CD = 3p, CE = 5p
CE^2=CD^2 + DE^2
(5p) ^2=(3p) ^2 + 8^2
25p^2 = 9p^2+ 64
25p^2- 9p^2 = 64
16p^2 =64
p^2=4
p=2
CD= 3P
CD = 6m
BD =BC + CD
BD = 11m
5) If sin B = ½, Show that 3cosB - 4COS^3B =0
Solution : We have,
sinB = ½ = perpendicular/hypotenuse
So let us draw a right triangle ABC, right angled at C such that Perpendicular = AC=1 unit, Hypotenuse = AB = 2 units.
Applying Pythagoras theorem in ABC, We obtain
AB^2 = BC^2 + AC^2
BC^2 = AB^2 - AC^2
BC^2 = 2^2 - 1^2
BC^2 = 4-1
BC^2 = 3
BC = 3
cos B = BC/ AB = 3 /2
3 cosB - 4 cos^2 B = 3x 3 /2- 4 (3 / 2)^3
= 33 / 2 - 4x 33 / 8
= 33 / 2 - 33 / 2
= 0
Hence proved
3 cos B - 4 cos^3 B = 0
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