JIPMAT 2023 (QA) - In ABC, B=90^, BC=5 \ cm, AC-AB=1 ~cm, then 1+ (c)1+ (c) is | PYQs + Solutions

JIPMAT 2023

Geometry

Trigonometry

Easy

In $\triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ}, \mathrm{BC}=5 \ \mathrm{cm}, \mathrm{AC}-\mathrm{AB}=1 \mathrm{~cm}$ , then $\frac{1+\sin (\mathrm{c})}{1+\cos (\mathrm{c})}$ is

\frac{5}{6}

\frac{6}{5}

2

\frac{25}{18}

Correct Option: 4

Knowing the Pythagorean triplets comes in handy:

5, 12, 13

(when we know the difference between two sides is

1

cm).

You can also find it using the Pythagorean theorem:

\newline

AB^2 + BC^2 = AC^2

Let's say

AB = x

, then

AC = x + 1

(since

AC - AB = 1

)

Using Pythagorean theorem:

\newline

x^2 + 25 = (x + 1)^2

\newline

x^2 + 25 = x^2 + 2x + 1

\newline

25 = 2x + 1

\newline

24 = 2x

\newline

x = 12

Therefore:

\newline

AB = 12

\newline

AC = 13

\newline

BC = 5

Find

\sin(C)

and

\cos(C)

\sin(C) = \frac{AB}{AC} = \frac{12}{13}

\cos(C) = \frac{BC}{AC} = \frac{5}{13}

Substituting into the expression:

\dfrac{1 + \sin(C)}{1 + \cos(C)} = \dfrac{1 + \frac{12}{13}}{1 + \frac{5}{13}}= \dfrac{13 +12}{13 + 5}= \dfrac{25}{18}