TRIGONOMETRY PRACTICE QUESTIONS: 21-30

Question 21.) If `sin` A = `\frac{3}{5}` and `sin` B = `\frac{5}{13}`. Now, if a right-angled triangle is formed whose one angle is A+B, then find the measure of the hypotenuse of this new right-angled triangle.

(A) 49 units

(B) 33 units

(C) 56 units

(D) 65 units

Answer. (D)

Solution.

Since it is given that,

`sin` A = `\frac{3}{5}`

&, `sin` B = `\frac{5}{13}`

Now if, `sin` A = `\frac{3}{5}`, then, `cos` A = `\frac{4}{5}`

& if, `sin` B = `\frac{5}{13}`, then, `cos` B = `\frac{12}{13}`

Now, we will use the formula of `sin` (A+B),

`sin` (A+B) = `sin` A `cos` B + `sin` B `cos` A

Now, putting the values of `sin` A, `cos` B, `sin` B, and `cos` A in the above formula, and doing further calculation, we will get

`sin` (A+B) = `\frac{56}{65}`

From here, it is clear that the measure of the hypotenuse is 65 units.

Question 22.) Consider the following equations,

`2^{2\cos\theta - 1}` + `2^{4 - 2\cos\theta}`  = 9

and,  `8^{\sin^{2}\phi}` + `8^{\cos^{2}\phi}` = 6

Now, if the value of `\tan`(`\theta` + `\phi`) is always negative, then the value of  `\phi` will always lie in which quadrant?

(A) Either Quadrant Ⅰ or Quadrant Ⅲ

(B) Either Quadrant Ⅰ or Quadrant Ⅳ

(C) Either Quadrant Ⅱ or Quadrant Ⅳ

(D) Either Quadrant Ⅲ or Quadrant Ⅳ

Answer. (A)

Solution.

Since it is given that,

`2^{2\cos\theta - 1}` + `2^{4 - 2\cos\theta}`  = 9

and,  `8^{\sin^{2}\phi}` + `8^{\cos^{2}\phi}` = 6

Let's consider the first equation,

`2^{2\cos\theta - 1}` + `2^{4 - 2\cos\theta}`  = 9

Let `2^{2\cos\theta}` = x

So, now the equation will become,

`\frac{x}{2}` + `\frac{16}{x}` = 9

On solving it, we get x = 2 or 16

Hence, `2^{2\cos\theta}` = 2 or 16

Now, only 2`\cos``\theta` = 1 can exist, not 4.

Hence, `\cos``\theta` = `\frac{1}{2}`

So, `\tan``\theta` = `\sqrt{3}`

Now, we will solve the 2nd equation,

`8^{\sin^{2}\phi}` + `8^{\cos^{2}\phi}` = 6

or, `8^{\sin^{2}\phi}` + `8^{1 - \sin^{2}\phi}` = 6

Let, `8^{\sin^{2}\phi}` = y

So, now the equation will become, y + `\frac{8}{y}` = 6

On solving it, we get y = 2 or 4

or, `8^{\sin^{2}\phi}` = 2 or 4

or, `2^{3\sin^{2}\phi}` = 2 or 4

or, 3`\sin^{2}``\phi` = 1 or 2

or, `\sin``\phi` = ±`\frac{1}{\sqrt{3}}` or ±`\sqrt{\frac{2}{3}}`

Hence, `\tan``\phi` = ±`\frac{1}{\sqrt{2}}` or ±`\sqrt{2}`

Now, as we know that, `tan`(A + B) = `frac{tanA + tanB}{1 - tanAtanB}`

Replacing A & B with `\theta` & `\phi` and further solving.

Here, we can see that `tan``\theta` ＞ `tan``\phi`

Hence, `tan``\theta` + `tan``\phi` ＞1 for any value of `tan``\phi`

So, for `tan`(`\theta` + `\phi`) to be always negative, the value of 1 -`tan``\theta``tan``\phi` has to be negative always.

And, it is possible only when `tan``\theta`.`tan``\phi` is always positive. It means the value of `tan``\phi` is always positive.

Hence, `\phi` will belong to either Quadrant Ⅰ or Quadrant Ⅲ.

Question 23.) What value of 𝛼 will satisfy the following equation,

|`\tan` 𝛼| = `\sec` 𝛼 - `\tan` 𝛼. Given that, 𝛼 ∈ [0, 2π].

(A) `\frac{π}{3}`

(B) `\frac{3π}{4}`

(C) `\frac{π}{6}`

(D) `\frac{2π}{3}`

Answer. (C)

Solution.

Since it is given that,

|`\tan` 𝛼| = `\sec` 𝛼 - `\tan` 𝛼

Here we will have two cases,

CASE 1, |`\tan` 𝛼| = -`\tan` 𝛼

By putting this value in the main equation, we will get, `\sec` 𝛼 = 0. Which is not possible.

CASE 2, |`\tan` 𝛼| = `\tan` 𝛼

By putting this value in the main equation, we will get, 

2`\tan` 𝛼 = `\sec` 𝛼

or,   `\sin` 𝛼 = `\frac{1}{2}`

Now, it is given that 𝛼 ∈ [0, 2π]

Hence, 𝛼 = either `\frac{π}{6}` or `\frac{5π}{6}`

But the value of `\tan` 𝛼 is positive. So, 𝛼 = `\frac{π}{6}`

Question 24.) If 7`tan^{2}``\phi` + `cot^{2}``\phi` = `\frac{179}{10}`. Then find that value of `cot^{2}``\phi` that satisfies 4`sin^{6}``\phi` + 5`cos^{6}``\phi` = `\frac{540}{343}`.

(A) `\frac{3}{7}`

(B) `\frac{25}{9}`

(C) `\frac{2}{5}`

(D) `\frac{136}{169}`

Answer. (C)

Solution.

Here we will first find the value of `cot^{2}``\phi` and then we will solve 4`sin^{6}``\phi` + 5`cos^{6}``\phi` for that value of `cot^{2}``\phi` to check it.

Since it is given that,

7`tan^{2}``\phi` + `cot^{2}``\phi` = `\frac{179}{10}`

Now, putting `tan``\phi` = `\frac{1}{\cot\phi}` and further solving we get,

10`cot^{4}``\phi` - 179`cot^{2}``\phi` + 70 = 0

10`cot^{4}``\phi` - 175`cot^{2}``\phi` - 4`cot^{2}``\phi` + 70 = 0

5`cot^{2}``\phi` (2`cot^{2}``\phi` - 35) - 2`cot^{2}``\phi` (2`cot^{2}``\phi` - 35) = 0

(2`cot^{2}``\phi` - 35).(5`cot^{2}``\phi` - 2) = 0

Hence, `cot^{2}``\phi` = `\frac{2}{5}` or `\frac{35}{2}`

Now, we have two values of `cot^{2}``\phi`. First, we will check the value of 4`sin^{6}``\phi` + 5`cos^{6}``\phi` for `cot^{2}``\phi` = `\frac{2}{5}`.

If, `cot^{2}``\phi` = `\frac{2}{5}`

Then, `sin^{2}``\phi` = `\frac{5}{7}`

And, `cos^{2}``\phi` = `\frac{2}{7}`

So, 4`sin^{6}``\phi` + 5`cos^{6}``\phi` = `\frac{540}{343}`

Hence, `cot^{2}``\phi` = `\frac{2}{5}` satisfies 4`sin^{6}``\phi` + 5`cos^{6}``\phi` = `\frac{540}{343}`.

Question 25.) If y = `2^{\tan^{2}x}` satisfies the following equation,

`y^{2}` - 3y +2 =0.

then for all values of x with in the range of [0, `\frac{π}{2}`] find the value of `frac{2cos x}{1 + sin x}`.

(A) 2`\sqrt{2}`

(B) `\frac{1}{\sqrt{3}}`

(C) 2`\sqrt{3}`

(D) 2`\sqrt{2}` + 1

Answer. (A)

Solution.

Here we will first solve the following equation for y,

`y^{2}` - 3y +2 =0

y = 1 or 2

Since it is given that,

y = `2^{\tan^{2}x}`

Comparing this value of y with the values evaluated, we get x = 0 or `\frac{π}{4}`

For x = 0, the value of `frac{2cos x}{1 + sin x}` = 2

And, for x = `\frac{π}{4}` the value of `frac{2cos x}{1 + sin x}` = 2(`\sqrt{2}`-1)

Now, 2 + 2(`\sqrt{2}`-1) = 2`\sqrt{2}`.

Question 26.) Find the value of, `\sin`(`\frac{1}{2}``tan^{-1}``\frac{5}{\sqrt{24}}`).

(A) `\frac{1}{4}`

(B) `\frac{1}{\sqrt{7}}`

(C) `\frac{1}{7}`

(D) `\frac{1}{2}`

Answer. (B)

Solution.

Since it is given that,

`\sin`(`\frac{1}{2}``tan^{-1}``\frac{5}{\sqrt{24}}`)

Let, `\frac{1}{2}``tan^{-1}``\frac{5}{\sqrt{24}}` = `\theta`

or, `\tan` 2`\theta` = `\frac{5}{\sqrt{24}}`

or, `\cos` 2`\theta` = `\frac{5}{7}`

or, 1 - 2`sin^{2}``\theta` = `\frac{5}{7}`

or, 2`sin^{2}``\theta` = 1 - `\frac{5}{7}` = `\frac{2}{7}`

or, `\sin` `\theta` = `\frac{1}{\sqrt{7}}`

Question 27.) A person is standing on the 8th floor of a 13-floor building A, where the lowest floor is marked as the ground floor. The person looks at the top of another building B, just opposite his building A, of the same height, and makes an angle of depression of 30°. If each floor covers a height of 5 meters, then at which floor of the building A should the man stand so that he makes an angle of depression of 60° from the top of the building B?

(A) 4th

(B) 3rd

(C) 2nd

(D) 1st

Answer. (D)

Solution.

The diagrammatic representation of the situation explained in the question is,

Now, in the given diagram, A & B are the two buildings with 13 floors each. X is the top-most floor of A, numbered as 12. The person is standing on the 8th floor of building A, which is marked as C, and makes an angle of depression of 30° to the top of building B. It means ∠XYC = 30°. Now, the person moves to one of the lower floors, which is marked as D, and from here he makes an angle of depression of 60° on top of building B. It means ∠XYD = 60°.

Now, we need to find which floor of building A is marked as D, and to do that, we need to determine its height above the ground floor.

Since it is given that each floor of the building covers 5 meters, the height of the building of 13 floors is

13 ✕ 5 = 65 meters

And the height of C from the ground floor is

9 ✕ 5 = 45 meters

Now, the length XC = 65 - 45 = 20 meters

Now let's consider the right-angled triangle XYC,

`\tan` ∠XYC = `\tan` 30° = `\frac{XC}{XY}`

or, `\frac{1}{\sqrt{3}}` = `\frac{20}{XY}`

Hence, XY = 20`\sqrt{3}` meters.

Now let's consider the right-angled triangle XYD,

`\tan` ∠XYD = `\tan` 60° = `\frac{XD}{XY}`

or, XD = XY ✕ `\tan` 60°

or, XD = 20`\sqrt{3}` ✕ `\sqrt{3}` = 60 meters

Hence, XD = 60 meters. So, its height from the ground floor = 65 - 6 = 5 meters.

Hence, D is the 1st floor of the building. So the man has to come to the 1st floor of building A to form an angle of depression of 60° from the top of building B.

Question 28.) A man is given a target to hit a car with a rocket from the top of a 60-meter-high tower. The man finds the car running away from the tower, 100 meters away. The man fires the rocket, and the rocket hits the car at a distance of 140 meters from the tower. Find the change in the angle of depression made by the car with the top of the tower when the man first looks at it and when the rocket hits it.

(A) `tan^{-1}``\frac{5}{22}`

(B) `tan^{-1}``\frac{3}{22}`

(C) `tan^{-1}``\frac{2}{\sqrt{11}}`

(D) `tan^{-1}``\frac{7}{38}`

Answer. (B)

Solution.

The diagrammatic representation of the situation explained in the question is,

In the given diagram, PQ is the tower of height 60 meters. The man hits the car from point P. R is the point where the man first spots the car, and its distance from the bottom of the tower PQ is, QR = 100 meters. S is the point where the man hits the car, and its distance from the bottom of the tower PQ is, QS = 140 meters.

Now, the angles of depression from points R & S on point P are ∠RPT & ∠SPT. We need to find the difference between these two angles.

Since PT ∥ QS. So, ∠RPT = ∠QRP & ∠SPT = ∠QSP. Hence, we need to find the difference between ∠QRP & ∠QSP.

Let's consider the triangle QRP,

`\tan` ∠QRP = `\frac{PQ}{QR}` = `\frac{60}{100}` = `\frac{3}{5}`

or, ∠QRP = `tan^{-1}`(`\frac{3}{5}`)

Similarly, let's consider the triangle QSP,

`\tan` ∠QSP = `\frac{PQ}{QS}` = `\frac{60}{140}` = `\frac{3}{7}`

or, ∠QSP = `tan^{-1}`(`\frac{3}{7}`)

Hence, the required difference between the angles of depression is given by,

∠QRP - ∠QSP = `tan^{-1}`(`\frac{3}{5}`) - `tan^{-1}`(`\frac{3}{7}`) = `tan^{-1}`(`\frac{3}{22}`).

Question 29.) A fighter jet flying 5 km above the ground enters the enemy's territory. But it gets spotted by the radar, and then an automatic system fires a missile at an angle of `tan^{-1}``\frac{1}{3}` towards it. But the fighter jet makes a sudden upward incline to protect itself from the missile. But the missile chases the jet and hits it at a height of 6 km above the ground level and at an inclined distance of 10 km from the automatic missile fire system. Find the inclined distance travelled by the fighter jet to protect itself from the missile.

(A) 8 km

(B) 2`\sqrt{13}` km

(C) `\sqrt{50}` km

(D) 1 km

Answer. (C)

Solution.

The diagrammatic representation of the situation explained in the question is,

In the given diagram, E is the point where the fighter jet arrives; then the missile from point A is launched to hit it, and the fighter jet takes the inclined path to protect itself. But the missile hits it at point C.

Now, it is given that the point E is 5 km above the ground level. So, DE = 5 km.

Similarly, point C is 6 km above the ground level. So, BC = 6 km.

And, this point C is 10 km away from the missile launcher. So, AC = 10 km.

Now, here we need to find the distance CE, and we will calculate it with the application of the Pythagoras theorem in the right-angled triangle CEF.

Now, CF = BC - BF. But BF = DE = 5 km, and BC = 6 km.

Hence, CF = 6 - 5 = 1 km.

And now we need to find EF.

From the diagram, it is clear that EF = BD, & BD = AD - AB.

Now, for the calculation of AD, we will consider the triangle ADE,

`\tan` ∠DAE = `\frac{DE}{AD}`

And, it is given that ∠DAE = `tan^{-1}`(`\frac{1}{3}`)

or, `\tan`[`tan^{-1}`(`\frac{1}{3}`)] = `\frac{5}{AD}`

Hence, AD = 15 km.

Now we will evaluate AB, and for that we will apply the Pythagoras theorem in triangle ABC,

`AB^{2}` + `BC^{2}` = `AC^{2}`

`AB^{2}` + `6^{2}` = `10^{2}`

On solving, we get AB = 8 km.

And, BD = 15 - 8 = 7 km.

Now, EF = BD = 7 km.

Let's consider triangle CEF,

`CE^{2}` = `FC^{2}` + `EF^{2}`

`CE^{2}` = `1^{2}` + `7^{2}`

On solving, we get CE = `\sqrt{50}` km.

Question 30.) If `\tan``theta`, and `\tan``\phi` are the roots of the following quadratic equation,

15`r^{2}` - 26r + 8 = 0

then what will be the measure of the hypotenuse of the triangle whose one angle will be `theta` - `\phi`.

(A) 5`\sqrt{29}` units.

(B) 15 units.

(C) 4`\sqrt{29}` units.

(D) 3`\sqrt{29}` units.

Answer. (A)

Solution.

Since the given equation is,

15`r^{2}` - 26r + 8 = 0

On solving it, we get its two roots that are as follows,

r = `\frac{4}{3}`, and `\frac{2}{5}`

Now, as per the question,

`\tan``theta` = `\frac{4}{3}`, and `\tan``\phi` = `\frac{2}{5}`

Now we will evaluate `\tan`(`theta` - `\phi`) by using the following formula,

`tan`(A - B) = `frac{tanA - tanB}{1 + tanAtanB}`

Now, putting the respective values and further solving, we get,

`\tan`(`theta` - `\phi`) = `\frac{14}{23}`

Now, let the measure of the hypotenuse be h; then,

h = `\sqrt{14^{2}+23^{2}}` = `\sqrt{725}` = 5`\sqrt{29}`

Hence, the measure of the hypotenuse = 5`\sqrt{29}` units.

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