CoCubes Trigonometry Questions and Answers Quiz-1

Let AB is the wall.
AC = 100 m = Ladder.
BC = distance between of ladder and wall = 60m.
Now, triangle ABC forms a right angle triangle.
AB² + BC² = AC
AB² = 100² - 60²
AB = 80m.
The face of the cube would be a square. Let that square has a length x.
There would be two right-angle triangles, ABC and QPC. With a common angle c.
Tan c = perpendicular/base
Tan c = 80/60 ------------- (1)
Tan c = x/(60 -x) ------- (2)
Comparing these two equation,
80/60 = x/(60 -x)
240 - 4x = 3x
7x = 240
x = 35m (Approx)
Thus, the side of the cube would be 35.

One of AB, AD and CD must have given.



So, the data is inadequate.

Let AB be the observer and CD be the tower.



Draw BE  CD.

Then, CE = AB = 1.6 m,

BE = AC = 20√3 m

DE/BE =  tan 30° = 1/√3 m

 CD = CE + DE = (1.6 + 20) m = 21.6 m.

Let AB be the tree and AC be its shadow.



Let ACB = .

Then, AC/AB =  √3  => cot θ = √3

θ = 30°.

Let AB be the tower.



Then, APB = 30° and AB = 100 m.

AB/AP = tan 30° = 1/√3

AP = (AB x √3) m

= 100√3 m

= (100 x 1.73) m

= 173 m

Answer: (B)

Explanation: =tan(60) + tan(180-60)

=tan(60)+(-tan(60)) [because in second quadrant tan is -ve]

=0

cos 2x = \frac{1 - tan^{2}x}{1 + tan^{2}x}

= \frac{1-\frac{25}{144}}{1 + \frac{25}{144}} = \left ( \frac{119}{144}\times \frac{144}{169} \right )

= \frac{119}{169}

Explanation: = 2sin²θ + 2cos²θ + cos²θ

=2(sin²θ + cos²θ) + cos²θ ; (by putting sin²θ + cos²=1)

2 + cos²θ ;(the minimum value of cos²θ=0)

= 2 + 0 = 2

Explanation: 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

Explanation:

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)