Test: Trigonometry- 2 - GMAT MCQ

Maximum value = √(a² + b²)
=13

Minimum value = √(a² – b²)
=-√2

tanA = cotB, 
tanA*tan B = 1
So, A +B = 90^o
(x+y)+(x-y) = 90^o, 2x = 90^o , x = 45^o
Tan (2x/3) = tan 30^o = 1/√3

tan(60) + tan(180-120)

=tan(60)+(-tan(60)) {because in second quadrant tan is -ve}

=0

2sin²θ + 2cos²θ + cos²θ
=2(sin²θ + cos²θ) + cos²θ ; (by putting sin²θ + cos²=1)
= 2 + cos²θ ;(the minimum value of cos²θ=0)
= 2 + 0 = 2

Let AB be the height of the tree and AC be the length of the shadow. We need to calculate the angle ACB where BC is the hypotenuse.
Given,
AC:AB = √3
cot θ = √3
θ = 30 degrees.

Let QR be the tower. Then,
QR = 100 m and angle QPR = 30 degrees.
We know, cot 30° = √3 = PQ/QR.
Therefore, PQ = 100*√3 = 173 m.

Let MN be the tower and man be standing at P (30° = angle MPN) and Q (60° = angle MQN). We are only given two angles and no sides of the triangles. Therefore, the data is inadequate.

Let the tower be PQ and the boat be at positions R and S when making angles of 45° and 30° respectively.
Given, PR = 60 m.
Now, PQ/PR = tan 45° = 1. So, PQ = PR = 60 m.
Again, PQ/PS = tan 30° = 1/√3. So, PS = 60 * √3 m = 103.92 m.
Distance covered in 5 seconds = 103.92 – 60 = 43.92 m.
Speed in kmph = (43.92/5) * (18/5) = 32 kmph (approximately)

Diagonal of Square = Diameter of Circle.

Let side of square be x.

From Pythogorous theorem.

Diagonal = √(2*x*x)

We know area of square = x * x = d ⇒ Diameter = Diagonal = √(2*d)

⇒  Radius = √(d/2)

Area of Circle = π * √(d/2) * √(d/2)

= 1/2 * π * d