Question for Practice Questions Level 2: Trigonometry - 1
Try yourself:Find the value of (cosθ cos(90 - θ) secθ) ÷ (tanθ sin (90 - θ)).
Explanation
(cosθ cos(90 - θ) secθ) ÷ (tanθ sin (90 - θ))
= cosθ sinθ secθ ÷ tanθ cosθ
= cosθ sinθ secθ ÷ cosθ sinθ secθ
= 1
Question for Practice Questions Level 2: Trigonometry - 1
Try yourself:If a = x sinθ + y cosθ and b = y sinθ - x cosθ, then y^{2} - a^{2} is equal to
Explanation
a = x sinθ+ y cosθ
b = y sinθ- x cosθ
y^{2} - a^{2} = y^{2} - (x sinθ+ y cosθ)^{2}
= y^{2} - x^{2} sin^{2}θ - y^{2} cos2θ - 2xy sinθ cosθ
= y^{2} sin^{2}θ - x^{2} sin^{2}θ - 2xy sinθ cosθ
= y^{2} sin^{2}θ + x^{2} cos^{2}θ - x^{2} - 2xy sinθ cosθ
= (y sinθ - x cosθ)^{2} - x^{2}
= b^{2} - x^{2}
Question for Practice Questions Level 2: Trigonometry - 1
Try yourself:If sin (π cos x) = cos (π sin x), then x equals
Explanation
Given: sin (π cos x) = cos (π sin x)
Question for Practice Questions Level 2: Trigonometry - 1
Try yourself:In the given figure, AB is the height of a tree. C and D are the two points in a straight line on the ground at b units and a units respectively away from the tree. What is the height of the tree (AB)?
Explanation
In ΔABC
AB/BC = tan 60°
AB = √3b units
In Δ ABD
AB/BD = tan 30°
AB = a/√3 units
Now AB × AB = √3b × a/√3 units^{2}
AB^{2} = ab units^{2}
AB = √ab units
Question for Practice Questions Level 2: Trigonometry - 1
Try yourself:A man on the top of a tower observes a car moving towards the base of the tower at an angle of depression α. Ten minutes later, the angle of depression of the car is found to be β. If the tangent of angle α. is equal to 1/√5, co-tangent of angle β is equal to 1/√5 and the car is moving with a uniform speed, find the total time taken by the car to reach the base of the tower.
Question for Practice Questions Level 2: Trigonometry - 1
Try yourself: Question for Practice Questions Level 2: Trigonometry - 1
Try yourself:If then the value of sin is Question for Practice Questions Level 2: Trigonometry - 1
Try yourself:The angles of elevation of the top of the tower, as observed from each of the points A, B and C on the ground, form a triangle at the same angle α. If R is the circum-radius of the triangle ABC, then the height of the tower is
Explanation
The tower makes equal angles at the vertices of the triangle.
Therefore, foot of the tower is at the circumcentre.
From ΔOAP, we have tanα = OP/OA
⇒ OP = OA tan α
⇒ OP = R tan α
Question for Practice Questions Level 2: Trigonometry - 1
Try yourself:If sin x + cos y = a and cos x + sin y = b, then what will be the value of tan (x - y)/2 ?
Question for Practice Questions Level 2: Trigonometry - 1
Try yourself:A vertical building and a tower are on the same ground level. From the top of the building, the angle of elevation of the top of the tower is 45° and the angle of depression of the foot of the tower is 60°. Find the height of the tower, if the height of the building is 30 m.