Answer
438.3k+ views
Hint: Here the properties of bisector are the only information which we use to reach the solution, which are as follows:
* Bisector of an angle A is defined to be the line cutting off A into equal halves which meets the
opposite side. In the below figure, AD is the bisector of angle A. Thus it divides angle A into 2 equal
halves and meets the opposite side BC.
* Bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle.
Implies point D divides BC into the ratio of sides containing angle A which are AC and AB. Thus
$\dfrac{{BD}}{{DC}} = \dfrac{{AC}}{{AB}}$
* Sine rule: Ratios of a side to corresponding opposite angles are equal. $\dfrac{{AB}}{{\sin C}} = \dfrac{{AC}}{{\sin B}} = \dfrac{{BC}}{{\sin A}}$
Complete step-by-step answer:
We start with bisector AD and develop its relation with the interior angles.
Sine rules on the triangle ABD gives the result: $\dfrac{{AD}}{{\sin B}} = \dfrac{{BD}}{{\sin (A/2)}}$
From the property of bisector and sine rule, $BD = \dfrac{c}{{b + c}}a$
$AD = \dfrac{{ca}}{{b + c}}\dfrac{{\sin B}}{{\sin (A/2)}}$
Multiplying numerator and denominator by $2\cos (A/2)$, for simplifying the equation to shorter terms.
$AD = \dfrac{{ca.\sin B.2\cos (A/2)}}{{(b + c)\sin (A/2).2\cos (A/2)}}$
Using the formula $\sin (2\theta ) = 2\sin \theta \cos \theta $ ,
$AD = 2.\dfrac{{ca.\sin B.\cos (A/2)}}{{(b + c)\sin A}}$
Using sine rule: $\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}}$ we get,
$
AD = 2.\dfrac{{cb\sin B\cos (A/2)}}{{(b + c)\sin B}} \\
= \dfrac{{2cb\cos (A/2)}}{{b + c}} \\
$
From the property of bisector and sine rule,
$AD = \dfrac{{2bc}}{{b + c}}\dfrac{{\cos (A/2)\sin (A/2)}}{{\sin (A/2)}}$
Simplifying,
$AD = \dfrac{{bc}}{{\sin (A/2)}}\dfrac{{\sin A}}{{\sin (A/2)}}$
Using $\sin A = \dfrac{a}{{2R}}$ , $\dfrac{1}{{\sin (A/2)}} = \csc (A/2)$ we get,
$AD = \dfrac{{abc}}{{2R(b + c)}}\csc (A/2)$
As $R = \dfrac{{abc}}{{4\Delta }}$
$AD = \dfrac{{abc.4\Delta }}{{2abc(b + c)}}\csc (A/2)$
Simplifying,
$AD = \dfrac{{2\Delta }}{{b + c}}\csc (A/2)$
So, the correct answer is “Option B” and “Option C”.
Note: Students are likely to make mistakes while substituting the sides using the sine formula, always keep in mind the angle opposite to the side comes in denominator along with the side in the numerator. Making the rearrangements of terms according to our motive is to be done properly for a quick solution. Multiplying and dividing by needed trigonometric terms can cut short the solution into simpler form lines.
* Bisector of an angle A is defined to be the line cutting off A into equal halves which meets the
opposite side. In the below figure, AD is the bisector of angle A. Thus it divides angle A into 2 equal
halves and meets the opposite side BC.
![seo images](https://www.vedantu.com/question-sets/9f8592a4-5e09-4086-b7ac-a77fb4ba5f345196138459157805434.png)
* Bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle.
Implies point D divides BC into the ratio of sides containing angle A which are AC and AB. Thus
$\dfrac{{BD}}{{DC}} = \dfrac{{AC}}{{AB}}$
* Sine rule: Ratios of a side to corresponding opposite angles are equal. $\dfrac{{AB}}{{\sin C}} = \dfrac{{AC}}{{\sin B}} = \dfrac{{BC}}{{\sin A}}$
Complete step-by-step answer:
We start with bisector AD and develop its relation with the interior angles.
Sine rules on the triangle ABD gives the result: $\dfrac{{AD}}{{\sin B}} = \dfrac{{BD}}{{\sin (A/2)}}$
From the property of bisector and sine rule, $BD = \dfrac{c}{{b + c}}a$
$AD = \dfrac{{ca}}{{b + c}}\dfrac{{\sin B}}{{\sin (A/2)}}$
Multiplying numerator and denominator by $2\cos (A/2)$, for simplifying the equation to shorter terms.
$AD = \dfrac{{ca.\sin B.2\cos (A/2)}}{{(b + c)\sin (A/2).2\cos (A/2)}}$
Using the formula $\sin (2\theta ) = 2\sin \theta \cos \theta $ ,
$AD = 2.\dfrac{{ca.\sin B.\cos (A/2)}}{{(b + c)\sin A}}$
Using sine rule: $\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}}$ we get,
$
AD = 2.\dfrac{{cb\sin B\cos (A/2)}}{{(b + c)\sin B}} \\
= \dfrac{{2cb\cos (A/2)}}{{b + c}} \\
$
From the property of bisector and sine rule,
$AD = \dfrac{{2bc}}{{b + c}}\dfrac{{\cos (A/2)\sin (A/2)}}{{\sin (A/2)}}$
Simplifying,
$AD = \dfrac{{bc}}{{\sin (A/2)}}\dfrac{{\sin A}}{{\sin (A/2)}}$
Using $\sin A = \dfrac{a}{{2R}}$ , $\dfrac{1}{{\sin (A/2)}} = \csc (A/2)$ we get,
$AD = \dfrac{{abc}}{{2R(b + c)}}\csc (A/2)$
As $R = \dfrac{{abc}}{{4\Delta }}$
$AD = \dfrac{{abc.4\Delta }}{{2abc(b + c)}}\csc (A/2)$
Simplifying,
$AD = \dfrac{{2\Delta }}{{b + c}}\csc (A/2)$
So, the correct answer is “Option B” and “Option C”.
Note: Students are likely to make mistakes while substituting the sides using the sine formula, always keep in mind the angle opposite to the side comes in denominator along with the side in the numerator. Making the rearrangements of terms according to our motive is to be done properly for a quick solution. Multiplying and dividing by needed trigonometric terms can cut short the solution into simpler form lines.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)