HSBC Trigonometry Questions and Answers Quiz-1

assumption is the cube is placed against the wall.

It should be 100m and 60m ..

let height =a=60, hypotenuse=c=100, so base=b=80

the formula is (a*b)/(a+b)=34.28

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(180-120)

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

=0

Explanation: 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.

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)