If $ A, B, C $, are the interior angles of a triangle $ A B C $, prove that
If $ \angle A=90^{\circ} $, then find the value of $ \tan \left(\frac{B+C}{2}\right) $.
Given:
\( A, B, C \), are the interior angles of a triangle \( A B C \).
\( \angle A=90^{\circ} \)
To do:
We have to find the value of \( \tan \left(\frac{B+C}{2}\right) \).
Solution:
We know that,
Sum of the angles in a triangle is $180^{\circ}$.
This implies,
$\angle A+\angle B+\angle C=180^{\circ}$
$\Rightarrow \frac{90^{\circ}+\angle B+\angle C}{2}=\frac{180^{\circ}}{2}$
$\Rightarrow \frac{90^{\circ}}{2}+ \frac{\angle B+\angle C}{2}=90^{\circ}$
$\Rightarrow \frac{\angle B+\angle C}{2}=90^{\circ}-45^{\circ}$
$\Rightarrow \frac{\angle B+\angle C}{2}=45^{\circ}$
Therefore,
$\tan \left(\frac{B+C}{2}\right)=\tan 45^{\circ}$
$=1$ (Since $\tan 45^{\circ}=1$)
The value of \( \tan \left(\frac{B+C}{2}\right) \) is $1$.
Related Articles
- If \( A, B, C \), are the interior angles of a triangle \( A B C \), prove that\( \tan \left(\frac{C+A}{2}\right)=\cot \frac{B}{2} \)
- If \( A, B, C \), are the interior angles of a triangle \( A B C \), prove that\( \sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2} \)
- If \( A, B, C \) are the interior angles of a \( \triangle A B C \), show that:\( \tan \frac{B+C}{2}=\cot \frac{A}{2} \)
- If \( A, B, C \) are the interior angles of a \( \triangle A B C \), show that:\( \cos \frac{B+C}{2}=\sin \frac{A}{2} \)
- If \( \triangle A B C \) is a right triangle such that \( \angle C=90^{\circ}, \angle A=45^{\circ} \) and \( B C=7 \) units. Find \( \angle B, A B \) and \( A C \).
- If $A, B$ and $C$ are interior angles of a triangle $ABC$, then show that: $sin\ (\frac{B+C}{2}) = cos\ \frac{A}{2}$
- Prove the following:\( \frac{\cos \left(90^{\circ}-A\right) \sin \left(90^{\circ}-A\right)}{\tan \left(90^{\circ}-A\right)}=\sin ^{2} A \)
- In a quadrilateral \( A B C D, \angle B=90^{\circ}, A D^{2}=A B^{2}+B C^{2}+C D^{2}, \) prove that $\angle A C D=90^o$.
- Prove that:\( \left(\frac{x^{a}}{x^{b}}\right)^{a^{2}+a b+b^{2}} \times\left(\frac{x^{b}}{x^{c}}\right)^{b^{2}+b c+c^{2}} \times\left(\frac{x^{c}}{x^{a}}\right)^{c^{2}+c a+a^{2}}=1 \)
- Prove that:\( \left(\frac{x^{a}}{x^{-b}}\right)^{a^{2}-a b+b^{2}} \times\left(\frac{x^{b}}{x^{-c}}\right)^{b^{2}-b c+c^{2}} \times\left(\frac{x^{c}}{x^{-a}}\right)^{c^{2}-c a+a^{2}}=1 \)
- In a right triangle \( A B C \), right angled at \( C \), if \( \angle B=60^{\circ} \) and \( A B=15 \) units. Find the remaining angles and sides.
- In right-angled triangle \( A B C \) in which \( \angle C=90 \), if \( D \) is the mid-point of \( B C \) prove that \( A B^{2}=4 A D^{2}-3 A C^{2} \).
- Show that:\( \left(\frac{x^{a^{2}+b^{2}}}{x^{a b}}\right)^{a+b}\left(\frac{x^{b^{2}+c^{2}}}{x^{b c}}\right)^{b+c}\left(\frac{x^{c^{2}+a^{2}}}{x^{a c}}\right)^{a+c}= x^{2\left(a^{3}+b^{3}+c^{3}\right)} \)
- The value of \( \left(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}\right) \) is(A) 0(B) 1(C) 2(D) \( \frac{1}{2} \)
- Prove that \( \sin \left(C+\frac{A+B}{2}\right)=\sin \frac{A+B}{2} \)
Kickstart Your Career
Get certified by completing the course
Get Started
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies