# Geometry - Trigonometry - Previous Year CAT/MBA Questions

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

**CAT 2019 QA Slot 1 | Geometry - Trigonometry**

The number of the real roots of the equation 2cos(x( x + 1)) = 2^{x} + 2^{–x} is

- A.
0

- B.
2

- C.
infinite

- D.
1

Answer: Option D

**Explanation** :

−2 ≤ 2cos(x( x + 1)) ≤ 2

∴ −2 ≤ 2^{x} + 2^{–x} ≤ 2

Let 2^{x} be a, so 2^{–x} is 1/a.

So, −2 ≤ a + (1/a) ≤ 2

∴ −2 ≤ (a^{2} + 1)/a ≤ 2

∴ −2a ≤ (a^{2} + 1) ≤ 2a

∴ (a^{2} + 1 + 2a) ≥ 0 ⇒ (a + 1)^{2} ≥ 0, so a ∈ R.

Also, a^{2} + 1 − 2a ≤ 0 ⇒ (a − 1)^{2} ≤ 0, so a = 1.

Hence a = 1.

So, 2^{x} = 1.

∴ x = 0.

So, there is only one real root.

Hence, option 4.

Workspace:

**CAT 2004 QA | Geometry - Trigonometry**

A father and his son are waiting at a bus stop in the evening. There is a lamp post behind them. The lamp post, the father and his son stand on the same straight line. The father observes that the shadows of his head and his son's head are incident at the same point on the ground. If the heights of the lamp post, the father and his son are 6 metres, 1.8 metres and 0.9 metres respectively, and the father is standing 2.1 metres away from the post, then how far (in metres) is the son standing from his father?

- A.
0.9

- B.
0.75

- C.
0.6

- D.
0.45

Answer: Option D

**Explanation** :

Let PA, FB and SC denote the lamp post, the father and the son respectively and let D denote the point where the shadows of the father’s and son’s heads are incident.

ΔBFD ~ ΔCSD

$\therefore \frac{BF}{CS}=\frac{FD}{SD}\phantom{\rule{0ex}{0ex}}\therefore \frac{1.8}{0.9}=\frac{FD}{SD}\phantom{\rule{0ex}{0ex}}\therefore \frac{FD}{SD}=2$

∴ FD = 2 × SD

Let SD = FS = x m

ΔAPD ~ ΔBFD

$\therefore \frac{AP}{BF}=\frac{PD}{FD}\phantom{\rule{0ex}{0ex}}\therefore \frac{6}{1.8}=\frac{2.1+2x}{2x}\phantom{\rule{0ex}{0ex}}\therefore \frac{20x}{3}=2.1+2x\phantom{\rule{0ex}{0ex}}\therefore \frac{14x}{3}=2.1$

∴ x = 0.45 m

∴ The son is standing 0.45 m away from the father.

Hence, option 4.

Workspace:

**CAT 2003 QA - Leaked | Geometry - Trigonometry**

A vertical tower OP stands at the centre O of a square ABCD. Let h and b denote the lengths OP and AB respectively. Suppose ∠APB = 60°, then the relationship between h and b can be expressed as

- A.
2

*b*^{2}=*h*^{2} - B.
2

*h*^{2}=*b*^{2} - C.
3

*b*^{2}= 2*h*^{2} - D.
3

*h*^{2}= 2*b*^{2}

Answer: Option B

**Explanation** :

OP = h and AB = b

Now, OA = $\frac{AC}{2}=\frac{b\sqrt{2}}{2}=\frac{b}{\sqrt{2}}$

OP is a perpendicular tower at the centre O of the square.

In ∆PAB, PA = PB

∴ ∠PAB = ∠PBA = ∠APB = 60°

∴ ∆PAB is an equilateral triangle.

∴ AP = b

In the right-angled ∆AOP, we have,

AP^{2} = OP^{2} + OA^{2}

∴ b^{2} = h^{2} + $\frac{{b}^{2}}{2}$

∴ 2h^{2} = b^{2}

Hence, option 2.

Workspace:

**CAT 2003 QA - Retake | Geometry - Trigonometry**

A car is being driven, in a straight line and at a uniform speed, towards the base of a vertical tower. The top of the tower is observed from the car and, in the process, it takes 10 minutes for the angle of elevation to change from 45° to 60°. After how much more time will this car reach the base of the tower?

- A.
$5\left(\sqrt{3}+1\right)$

- B.
$6\left(\sqrt{3}+1\right)$

- C.
$7\left(\sqrt{3}-1\right)$

- D.
$8\left(\sqrt{3}-2\right)$

Answer: Option A

**Explanation** :

Let x be the distance from the later position of the car and the tower (i.e. when the angle of elevation was 60°).

Since the triangle formed (i.e. ∆ABD) is a 30°-60°-90° triangle, we have,

height of the tower, h = x$\sqrt{3}$

Now, since the triangle formed by the initial position of the car (i.e. ∆ABC) is an isosceles triangle, AB = BC

i.e. BC = $x\sqrt{3}$

∴ DC = x$\sqrt{3}$ - x = x($\sqrt{3}$ - 1)

Time taken to travel distance DC is 10 minutes, thus,

Speed $s=\frac{x(\sqrt{3}-1)}{10}$

Time taken to travel distance x = $\frac{x}{{\displaystyle \frac{x(\sqrt{3}-1)}{10}}}=\frac{10}{\sqrt{3}-1}=5(\sqrt{3}+1)$

Hence, option 1.

Workspace:

**CAT 2001 QA | Geometry - Trigonometry**

A ladder leans against a vertical wall. The top of the ladder is 8 m above the ground. When the bottom of the ladder is moved 2 m farther away from the wall, the top of the ladder rests against the foot of the wall. What is the length of the ladder?

- A.
10 m

- B.
15 m

- C.
20 m

- D.
17 m

Answer: Option D

**Explanation** :

Let the foot of the ladder be x metres away from the foot of the wall.

Now, the length of the ladder will be = x + 2

∴ (x + 2)^{2} = x^{2} + 8

On solving, we get, x = 15

∴ Length of ladder = 15 + 2 = 17 m

Hence, option 4.

Workspace:

**CAT 1999 QA | Geometry - Trigonometry**

**Directions: **Each question is followed by two statements I and II. Mark:

1. if the question can be answered by any one of the statements alone, but cannot be answered by using the other statement alone.

2. if the question can be answered by using either statement alone.

3. if the question can be answered by using both the statements together, but cannot be answered by using either statement alone.

4. if the question cannot be answered even by using both the statements together.

A line graph on a graph sheet shows the revenue for each year from 1990 through 1998 by points and joins the successive points by straight-line segments. The point for revenue of 1990 is labelled A, that for 1991 as B, and that for 1992 as C. What is the ratio of growth in revenue between 1991-92 and 1990-91?

I. The angle between AB and X-axis when measured with a protractor is 40°, and the angle between CB and X-axis is 80°.

II. The scale of Y-axis is 1 cm = Rs. 100

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

**Explanation** :

Ratio of revenues = $\frac{RQ}{QP}$

Since in a line graph, the years are uniformly spaced

$\Rightarrow \frac{RQ}{QP}=\frac{\mathrm{tan}80\xb0}{\mathrm{tan}40\xb0}$

So the ratio can be determined from statement I alone.

Statement II is immaterial because we intend to find the ratio and not absolute figures.

Workspace:

## Feedback

**Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing
us your valuable feedback about Apti4All and how it can be improved.**