Let A, B and C are the angles of a triangle and (A)/(2)=(1)/(3), (B)/(2)=(2)/(3) . Then, (C)/(2) is equal to

Correct answer: A. (7)/(9) aligned A+B+C=180^ ^ (Angle sum property of a triangle) & A+B=180^ ^ -C & (A+B)/(2)=90^ ^ -(C)/(2) & ((A)/(2)+(B)/(2))= (90^ ^ -(C)/(2)) & ( A / 2+ B / 2)/(1- A / 2 B / 2)= C / 2 & ((1)/(3)+(2)/(3))/(1-(1)/(3) × (2)/(3))= C / 2 (1)/(1-(2)/(9))= C / 2 & (9)/(7)= C / 2 C / 2=7 / 9 aligned )

BITSAT 2024

Exam: BITSAT
Year: 2024
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

BITSAT 2024 · Mathematics · Trigonometric Ratios & Identities

Let A, B and C are the angles of a triangle and (A)/(2)=(1)/(3),...

Let

A, B

and

C

are the angles of a triangle and

\tan \frac{A}{2}=\frac{1}{3}, \tan \frac{B}{2}=\frac{2}{3}

. Then,

\tan \frac{C}{2}

is equal to

Options

A $\frac{7}{9}$
B $\frac{2}{9}$
C $\frac{1}{3}$
D $\frac{2}{3}$

Correct answer

A. $\frac{7}{9}$

Step-by-step solution

$\begin{aligned} \mathrm{A}+\mathrm{B}+\mathrm{C}=180^{{}^{\circ}} \text{ (Angle sum property of a triangle) }\\ & \Rightarrow \mathrm{A}+\mathrm{B}=180^{{}^{\circ}}-\mathrm{C}\\ & \Rightarrow \frac{A+B}{2}=90^{{}^{\circ}}-\frac{C}{2}\\ & \Rightarrow \tan \left(\frac{A}{2}+\frac{B}{2}\right)=\tan \left(90^{{}^{\circ}}-\frac{C}{2}\right)\\ & \Rightarrow \frac{\tan A / 2+\tan B / 2}{1-\tan A / 2 \tan B / 2}=\cot C / 2 \\ & \Rightarrow \frac{\frac{1}{3}+\frac{2}{3}}{1-\frac{1}{3} \times \frac{2}{3}}=\cot C / 2 \Rightarrow \frac{1}{1-\frac{2}{9}}=\cot C / 2 \\ & \Rightarrow \frac{9}{7}=\cot C / 2 \Rightarrow \tan C / 2=7 / 9\end{aligned}

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