# Geometry - Trigonometry - Previous Year CAT/MBA Questions

You can practice all previous year OMET questions from the topic Geometry - Trigonometry. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.

**IIFT 2019 QA | Geometry - Trigonometry**

A man standing on the line joining the two poles finds that the top of the poles make an angle of elevation of 60° and 45° respectively. After walking for sometime towards the other pole, the angles change to 30° and 60° respectively. The ratio of the height of the poles is:

- A.
[(√3) − 1]/2

- B.
[(√3) + 1]/2

- C.
[(√3) − 1]/4

- D.
[(√3) + 1]/4

Answer: Option A

**Explanation** :

Consider the image below where A and B are the initial and final position of the man. PQ and LM are the two poles of heights a and b respectively.

From ∆PQA, QA = a/Tan60 = a/√3. From ∆ALM, AM = b/Tan45 = b.

From ∆PQB, QB = a/Tan30 = a√3. From ∆BLM, BM = b/Tan60 = b/√3.

∴ [a/√3] + b = a√3 + [b/√3]

Solving we get,

a/b = [(√3) − 1]/2

Hence option (a).

Workspace:

**XAT 2018 QADI | Geometry - Trigonometry**

A boat, stationed at the North of a lighthouse, is making an angle of 30° with the top of the lighthouse. Simultaneously, another boat, stationed at the East of the same lighthouse, is making an angle of 45° with the top of the lighthouse. What will be the shortest distance between these two boats? The height of the lighthouse is 300 feet. Assume both the boats are of negligible dimensions.

- A.
300 feet

- B.
600√3 feet

- C.
300√3 feet

- D.
600 feet

- E.
None of the above

Answer: Option D

**Explanation** :

Let LM be the lighthouse and B_{1} and B_{2} be the positions of the two boats.

In ∆LMB_{1},

Tan 30° = LM/MB_{1}

⇒ MB_{1} = LM√3 = 300√3

Also, in ∆LMB_{2},

Tan 45° = LM/MB_{2}

⇒ MB_{2} = LM = 300

In ∆MB_{1}B_{2},

(B_{1}B_{2})^{2} = (MB_{1})^{2} + (MB_{2})^{2}

⇒ (B_{1}B_{2})^{2} = (300√3)^{2} + (300)^{2}

⇒ (B_{1}B_{2})^{2} = 300^{2} × [(√3)^{2} + (1)^{2}]

⇒ (B_{1}B_{2})^{2} = 300^{2} × 4

⇒ B_{1}B_{2} = 300 × 2 = 600

Hence, option (d).

Workspace:

**IIFT 2018 QA | Geometry - Trigonometry**

At the foot of the mountain, the angle of elevation of the summit at the top of the mountain is 45⁰. After ascending 100 metres, at a slope of 30⁰ up the mountain towards the summit, the angle of elevation of the summit is 60⁰. Find the height of the summit.

- A.
$50\left(\sqrt{3}+1\right)$ metres

- B.
$50\left(\sqrt{5}+1\right)$ metres

- C.
$50\left(\sqrt{3}+2\right)$ metres

- D.
$50\sqrt{3}$ metres

Answer: Option A

**Explanation** :

Let height of the mountain = x

From the figure, AD = distance travelled by man = 100 m

Also, DE = AD × sin 30° = 100 × 0.5 = 50 m

Also, AE = AD × cos 30° = 100 × (√3/2) = 50√3 m

Now, ∠CAB = 45°

Hence, ABC is a 45-45-90 triangle and AB = BC = x

Also, EB = AB – AE = (x − 50√3) m

∴ DF = EB = (x − 50√3) m

Similarly, DE = FB = 50 m

∴ CF = BC – FB = (x – 50) m

Now, in ∆CDF, tan 60° = CF/DF

∴ √3 = (x – 50)/(x − 50√3)

∴ √3x – 50(3) = x − 50

∴ (√3 – 1)x = 150 – 50 = 100

∴ x = 100/((√3 – 1) = 50(√3 + 1) [Rationalising]

Hence, option 1.

Workspace:

**XAT 2017 QADI | Geometry - Trigonometry**

If 5° ≤ x° ≤ 15°, then the value of sin 30° + cos x° - sin x° will be:

- A.
Between -1 and -0.5 inclusive

- B.
Between -0.5 and 0 inclusive

- C.
Between 0 and 0.5 inclusive

- D.
Between 0.5 and 1 inclusive

- E.
None of the above

Answer: Option E

**Explanation** :

sin 30° + cos x° – sin x° where 5° ≤ x° ≤ 15°.

For the given range, cos x > sin x

So, cos x° – sin x° > 0

Also, sin 30° = 0.5

sin 30° + cos x° – sin x° > 0.5

So, first four options are eliminated.

Hence, option (e).

Workspace:

**IIFT 2017 QA | Geometry - Trigonometry**

A flag pole on the top of a mall building is 75 m high. The height of the mall building is 325 m. To an observer at height of 400 m, the mall building and the pole subtend an equal angle Ө. If the horizontal distance of the observer from the pole is x, what is the value of x?

- A.
20√10 m

- B.
30√10 m

- C.
25√5 m

- D.
None

Answer: Option B

**Explanation** :

Consider the layout of the mall, flag pole and oberver as shown below.

Let the horizontal distance from the oberver to the pole be *d* m.

**Considering pole: ** tan *θ* = 75/*d* ... (i)

**Considering mall building:** tan 2*θ* = 400/*d* ... (ii)

∴ (2tan*θ*) / (1 − tan^{2}*θ*) = 400/*d*

∴ 2(75/*d*) = (400/*d*) × [1 − (75/*d*)^{2}]

∴ 1 − (5625/*d*^{2}) = 3/8

∴ (5625/*d*^{2}) = 5/8

∴ *d*^{2} = 5625 × (8/5) = 9000

∴ *d* = 900 × 10 = 30√10 m

Hence, option 2.

Workspace:

**XAT 2016 QADI | Geometry - Trigonometry**

A person standing on the ground at point A saw an object at point B on the ground at a distance of 600 meters. The object started flying towards him at an angle of 30° with the ground. The person saw the object for the second time at point C flying at 30° angle with him. At point C, the object changed direction and continued flying upwards. The person saw the object for the third time when the object was directly above him. The object was flying at a constant speed of 10 kmph.

Find the angle at which the object was flying after the person saw it for the second time. You may use additional statement(s) if required.

Statement I: After changing direction the object took 3 more minutes than it had taken before.

Statement II: After changing direction the object travelled an additional 200√3 meters.

Which of the following is the correct option?

- A.
Statement I alone is sufficient to find the angle but statement II is not.

- B.
Statement II alone is sufficient to find the angle but statement I is not.

- C.
Statement I and Statement II are consistent with each other.

- D.
Statement I and Statement II are inconsistent with each other.

- E.
Neither Statement I nor Statement II is sufficient to find the angle.

Answer: Option D

**Explanation** :

From the given data,

m∠CAB = 30°; m∠CBA = 30° and AB = 600 m

∴ BC = AC = 200√3 m … [By sine rule]

Statement I: After changing the direction the object took 3 more minutes than it had taken before.

The object travels 200√3 m from B to C at 10 km/hr

Thus, in 3 minutes it can travel 500 m. Hence, the object travels a total of 500 + 200√3m from C.

Thus, we know the hypotenuse CD by which we can find out the angle.

Statement II: After changing directions, the object travels 200√3 m.

Since, the object travels the same distance as before, this can only happen if the object stays on the course as before without changing any direction.

Thus, we can clearly see that the two angles from the statements are inconsistent with each other.

Hence, option (d).

Workspace:

**XAT 2015 QA | Geometry - Trigonometry**

A person is standing at a distance of 1800 meters facing a giant clock at the top of a tower. At 5.00 p.m., he can see the tip of the minute hand of the clock at 30 degree elevation from his eye-level. Immediately, the person starts walking towards the tower. At 5.10 pm., the person noticed that the tip of the minute hand made an angle of 60 degrees with respect to his eye-level. Using three-dimensional vision, find the speed at which the person is walking. The length of the minutes hand is 200√3 meters (√3 = 1.732).

- A.
7.2 km/hour

- B.
7.5 km/hour

- C.
7.8 km/hour

- D.
8.4 km/hour

- E.
None of the above

Answer: Option D

**Explanation** :

At 5.00 p.m., position of the person be P

PB = 1800 (Given)

m∠DPB = 30°

⇒ DB = 1800 tan 30° = 600√3

At 5.10 p.m. the minute hand of the clock moves by 60°

DC = CF = 200√3 m (Given)

∆FEC is 30° - 60° - 90° triangle.

So, EC = 100√3 m and EF = 300 m

DE = DC – EC = 200√3 – 100√3 = 100√3 m

FG = DB – DE = 600√3 – 100√3 = 500√3 m

In ∆AFG, m∠FAG = 60° and FG = 500√3 m

By theorem of 30°-60°-90° triangle,

AG = 500 m

BG = EF = 300 m

In ∆ABG, AG = 500 m and BG = 300 m ⇒ AB = 400 m

PA = PB – AB = 1800 – 400 = 1400 m = 1.4 km

Time taken = 10 minutes = (1/6) hours

Speed = 1.4 × 6 = 8.4 km/hr

Hence, option 4.

Workspace:

**IIFT 2014 QA | Geometry - Trigonometry**

The value of log_{7} log_{7} $\sqrt{7\sqrt{7\sqrt{7}}}$ is equal to:

- A.
7

- B.
log

_{7}2 - C.
1 – 3 log27

- D.
1 – 3 log

_{7}2

Answer: Option D

**Explanation** :

log_{7} log_{7} $\sqrt{7\sqrt{7\sqrt{7}}}$ log_{7} log_{7} ${7}^{\frac{7}{8}}$

= log_{7} $\frac{7}{8}$ = 1 - 3log_{7} 2

Hence, option 4.

Workspace:

**IIFT 2010 QA | Geometry - Trigonometry**

If each α, β, γ is a positive acute angle such that

sin(α + β + γ) = $\frac{1}{\sqrt{2}}$, cosec(β + γ - α) = $\frac{2}{\sqrt{3}}$ and tan(γ + α - β) = $\frac{1}{\sqrt{3}}$

What are the values of α, β, γ?

- A.
$\left(37\frac{1}{2},52\frac{1}{2},45\right)$

- B.
(37, 53, 45)

- C.
$\left(45,37\frac{1}{2},52\frac{1}{2}\right)$

- D.
$\left(34\frac{1}{2},55\frac{1}{2},45\right)$

Answer: Option A

**Explanation** :

We evaluate options:

Option (1) Let α = 37.5, β = 52.5, γ = 45.

∴ α + β – γ = 45

∴ sin (α + β – γ) = $\frac{1}{\sqrt{2}}$

β + γ – α = 60

cosec(β + γ – α) = $\frac{2}{\sqrt{3}}$

γ + α – β = 30

tan(γ + α – β) = $\frac{1}{\sqrt{3}}$

Hence, option 1.

Workspace:

**IIFT 2010 QA | Geometry - Trigonometry**

The minimum value of 3^{sinx}* *+ 3^{cosx} is

- A.
2

- B.
2$\left({3}^{-\frac{1}{\sqrt{2}}}\right)$

- C.
${3}^{1-\frac{1}{\sqrt{2}}}$

- D.
None of these

Answer: Option B

**Explanation** :

*f*(*x*) = 3^{sinx} + 3^{cos x}

Using the inequality, AM ≥ GM

*f*(*x*) will have a minimum

where 3 ^{sin x}* *= 3 ^{cos x}

∴ sin *x *= cos *x*

∴ sin *x *= cos *x = -$\frac{1}{\sqrt{2}}$*

f(x)_{min} = ${3}^{-\frac{1}{\sqrt{2}}}$ + ${3}^{-\frac{1}{\sqrt{2}}}$

= 2$\left({3}^{-\frac{1}{\sqrt{2}}}\right)$

Hence, option 2.

Workspace:

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