# Geometry - Trigonometry - Previous Year CAT/MBA Questions

The best way to prepare for Geometry - Trigonometry is by going through the previous year **Geometry - Trigonometry XAT questions**.
Here we bring you all previous year Geometry - Trigonometry XAT questions along with detailed solutions.

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**XAT 2023 QADI | Geometry - Trigonometry XAT Question**

Find the value of:

$\frac{{\mathrm{sin}}^{6}15\xb0+{\mathrm{sin}}^{6}75\xb0+6{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}{{\mathrm{sin}}^{4}15\xb0+{\mathrm{sin}}^{4}75\xb0+5{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}$

- (a)
sin 15° + sin 75°

- (b)
6/5

- (c)
1

- (d)
sin 15° cos 15°

- (e)
None of the above

Answer: Option C

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**Text Explanation** :

$\frac{{\mathrm{sin}}^{6}15\xb0+{\mathrm{sin}}^{6}75\xb0+6{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}{{\mathrm{sin}}^{4}15\xb0+{\mathrm{sin}}^{4}75\xb0+5{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}$

Let sin^{2}15° = a and sin^{2}75° = b

∴ $\frac{{\mathrm{sin}}^{6}15\xb0+{\mathrm{sin}}^{6}75\xb0+6{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}{{\mathrm{sin}}^{4}15\xb0+{\mathrm{sin}}^{4}75\xb0+5{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}$ = $\frac{{\mathrm{a}}^{3}+{\mathrm{b}}^{3}+6\mathrm{ab}}{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}+5\mathrm{ab}}$

Now, a^{3} + b^{3} = (a + b)^{3} - 3ab(a + b), and

a^{2} + b^{2} = (a + b)^{2} - 2ab

∴ $\frac{{\mathrm{sin}}^{6}15\xb0+{\mathrm{sin}}^{6}75\xb0+6{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}{{\mathrm{sin}}^{4}15\xb0+{\mathrm{sin}}^{4}75\xb0+5{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}$ = $\frac{{(\mathrm{a}+\mathrm{b})}^{3}-3\mathrm{ab}(\mathrm{a}+\mathrm{b})+6\mathrm{ab}}{{(\mathrm{a}+\mathrm{b})}^{2}-2\mathrm{ab}+5\mathrm{ab}}$

Now, a + b = sin^{2}15° + sin^{2}75° = sin^{2}15° + cos^{2}15° = 1 [β΅ Sinπ = Cos(90 - π)]

∴ $\frac{{\mathrm{sin}}^{6}15\xb0+{\mathrm{sin}}^{6}75\xb0+6{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}{{\mathrm{sin}}^{4}15\xb0+{\mathrm{sin}}^{4}75\xb0+5{\mathrm{sin}}^{2}15\xb0{\mathrm{sin}}^{2}75\xb0}$ = $\frac{1-3\mathrm{ab}+6\mathrm{ab}}{1-2\mathrm{ab}+5\mathrm{ab}}$ = $\frac{1+3\mathrm{ab}}{1+3\mathrm{ab}}$ = 1

Hence, option (c).

Workspace:

**XAT 2022 QADI | Geometry - Trigonometry XAT Question**

A tall tower has its base at point K. Three points A, B and C are located at distances of 4 metres, 8 metres and 16 metres respectively from K. The angles of elevation of the top of the tower from A and C are complementary. What is the angle of elevation (in degrees) of the tower’s top from B?

- (a)
60

- (b)
30

- (c)
45

- (d)
We need more information to solve this.

- (e)
15

Answer: Option C

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**Text Explanation** :

βββββββ

Given the distances are :

AE = 4 meters , EB = 8 meters and EC = 16 meters.

Considering the length of ED = K.

Given the angles DAE and angle DCE are complementary.

Hence the angles are A and 90 - A.

Tan(90 - A) = Cot A

tan DAE = $\frac{k}{4}$ and tan DCE = tan $\frac{1}{DAE}$ = $\frac{k}{16}$

Hence $\frac{k}{16}$ = $\frac{4}{k}$

k = 8 meters.

The angle DBE is given by

tan DBE = $\frac{k}{8}$ = 1

Hence the angle is equal to 45 degrees.

Workspace:

**XAT 2018 QADI | Geometry - Trigonometry XAT Question**

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

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**Text Explanation** :

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

In βLMB_{1},

Tan 30° = LM/MB_{1}

⇒ MB_{1} = LM√3 = 300√3

Also, in βLMB_{2},

Tan 45° = LM/MB_{2}

⇒ MB_{2} = LM = 300

In βMB_{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:

**XAT 2017 QADI | Geometry - Trigonometry XAT Question**

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

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**Text 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:

**XAT 2016 QADI | Geometry - Trigonometry XAT Question**

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.

βββββββ

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

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**Text Explanation** :

From the given data,

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

∴ BC = AC = 200√3 m … [By sine rule]

βββββββ

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 XAT Question**

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

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**Text Explanation** :

βββββββ

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°

βββββββ

DC = CF = 200√3 m (Given)

βFEC 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 βAFG, m∠FAG = 60° and FG = 500√3 m

By theorem of 30°-60°-90° triangle,

AG = 500 m

BG = EF = 300 m

In βABG, 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 (d).

Workspace:

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