Question for Practice Questions Level 1: Trigonometry - 2
Try yourself:From a 60 m high building, the angles of depression of two cars on the opposite ends of the building are observed to be 60° and 30°. Find the distance between the cars, if the line joining them passes through the foot of the building.
Explanation
Suppose OP is the building, and A and B are the two cars. Then, OP = 60, PAO = 60° and PBO = 30°
From the right triangle OBP, tan 30° = 60/OB
OB = 60 √m ...(i)
Also, from the right triangle AOP, tan 60° = 60/OA
OA = 60/√3 ....(ii)
Thus, from (i) and (ii),
Question for Practice Questions Level 1: Trigonometry - 2
Try yourself:If 10 sinθ = 6, then tanθ + cotθ =
Explanation
If 10 sinθ = 6,
sinθ = 6/10 = 3/5
Question for Practice Questions Level 1: Trigonometry - 2
Try yourself:A right triangle PQR is right-angled at Q. If tan R = 1/√3, then which of the following is the value of sin R?
Explanation
In a right-angled triangle PQR with ∠Q = 90°:
sin R = p/h , cos R = b/h, tan R = p/b, cosec R = h/p , sec R = h/b and cot R = b/p.
Here, p stands for perpendicular, b stands for base and h stands for hypotenuse.
Now, in ΔPQR, tan R = 1/√3 ……… (1)
Also, tan R = p/b……… (2)
On comparing (1) and (2), we get
p = 1 and b = √3
Using Pythagoras theorem:
h^{2} = p^{2} + b^{2}
h =
h = 2
Thus, p = 1 and h = 2
So, sin R = p/h = 1/2.
Question for Practice Questions Level 1: Trigonometry - 2
Try yourself:If sin 30° = 1/2 , the value of expression tan 30° - sin 15° cos 15° is
Question for Practice Questions Level 1: Trigonometry - 2
Try yourself:In a right triangle ABC, right-angled at B, cot^{2} A = 9/16. What is the value of cos A?
Explanation
In right ΔABC,
∠B = 90°, cot^{2} A = 9/16
Taking square root on both sides, we get
cot A = 3/4… (1)
From (1) and (2), we get
b/p = 3/4
⇒ b = 3k and p = 4k
Now, using Pythagoras theorem,
AC^{2} = AB^{2} + BC^{2}
h^{2} = p^{2} + b^{2}
= (4k)^{2} + (3k)^{2}
= 16k^{2} + 9k^{2}
= 25k^{2}
h = 5k
cos A = b/h = 3/5
Question for Practice Questions Level 1: Trigonometry - 2
Try yourself:Find the value of sec 30° cot 30° tan 30° cosec 30°.
Explanation
Therefore, sec 30° = 2/√3 = cosec 30° = 2, tan 30° = 1/√3 and cot 30° = √3
Hence, sec 30° tan 30° cot 30° cosec 30°
Question for Practice Questions Level 1: Trigonometry - 2
Try yourself:If m = 1 + tan^{2}θ and n = 1 + cot^{2}θ, then m/n equals
Explanation
1 + tan^{2}θ = sec^{2}θ,
1 + cot^{2}θ = cosec^{2}θ
and cotθ = 1/tanθ
Given:
m = 1 + tan^{2}θ
Question for Practice Questions Level 1: Trigonometry - 2
Try yourself:If sinθ = 3/5, then find the value of (cosθ + tanθ + cotθ).
Explanation
When sinθ = 3/5, we can draw a triangle as shown in the figure.
sinθ = CB/AC = 3/5 (given)
cosθ = AB/AC = 4/5
tanθ = BC/AB = 3/4
cotθ = AB/BC = 4/3
cosθ + tanθ + cotθ = (4/5) + (3/4) + (4/3)
= (48 + 45 + 80)/60 = 173/60
Question for Practice Questions Level 1: Trigonometry - 2
Try yourself: Question for Practice Questions Level 1: Trigonometry - 2
Try yourself: then what is the value of A? Explanation
or sec A = 2
or sec A = sec 60°
or A = 60°
Therefore, option (1) is correct.
Question for Practice Questions Level 1: Trigonometry - 2
Try yourself:From the top of a 10 m high building, the angle of elevation of the top of a tower is 45° and the angle of depression of its foot is 60°. How will you represent this situation diagrammatically?
Explanation
The observer is at point 'D'.
Question for Practice Questions Level 1: Trigonometry - 2
Try yourself:The reduced form of cos^{6} x + sin^{6} x + 3 cos^{2} x.sin^{2} x is _______________.
Explanation
We know that,
cos^{2} x + sin^{2} x = 1
Taking the third power,
(cos^{2} x + sin^{2} x)^{3} = cos^{6} x + sin^{6} x + 3 cos^{2} x sin^{4}x + 3 sin^{2} x cos^{4} x = 1
i.e. cos^{6} x + sin^{6} x + 3 cos^{2} x sin^{2} x (sin^{2} x + cos^{2}x) = 1
cos^{6} x + sin^{6} x + 3 cos^{2} x sin^{2} x = 1
Question for Practice Questions Level 1: Trigonometry - 2
Try yourself:tan^{2}θ (1 + cot^{2}θ) (cosec^{2}θ + sec^{2}θ)
Explanation
tan^{2}θ (1 + cot^{2}θ) (cosec^{2}θ + sec^{2}θ)
Question for Practice Questions Level 1: Trigonometry - 2
Try yourself:If in a right triangle, tan A = 3/4 , then find the value of
Explanation
tan A = 3/4
⇒ sin A = 3/5 and cos A = 4/5
Question for Practice Questions Level 1: Trigonometry - 2
Try yourself:From two points A and B on the opposite sides of a tower, the angles of elevation to the top of the tower are 45° and 30°, respectively. If the height of the tower is 120 m, then find the distance between A and B, corrected to two decimal places.
Explanation
Let the length of the tower be CD.
CD = 120 m
∠DAC = 45° and DBC = 30°
In ΔADC,
DC/AC = tan 45°
120/AC = 1
AC = 120 m
CB = 120 × 1.732
CB = 207.84 m
Thus, distance between points A and B
= AC + CB = 120 + 207.84 = 327.84 m