Question

In a triangle angle ABC =90 and angle BAC= 60 the bisector of angle BAC meets BC at D, then ratio between BD and DC is

Answer 1

In a triangle ABC, AD is the bisector of angle BAC, if AB=6cm, AC=5cm and BD=3, then DC=?

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3 ANSWERS



Faizu Shahab M, studied at Calicut University Institute of Engineering and Technology

Answered Oct 17, 2017 · Author has 195answers and 242.5k answer views

Let the given ∆ABC be as drawn below:



Here, we can see that  is bisected by AD where AD touches BC at D. where, . We also have the values of adjacent sides and .

These satisfy the requirements of angle bisector theorem. Then, by the theorem, we have:

Happy math!!

3.9k Views · View 9 Upvoters · Answer requested by Prafful Sharma

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Girija Warrier, Passionate about Maths & Music...

Answered Oct 15, 2017 · Author has 1.9kanswers and 1.6m answer views

By Angle bisector theorem: which states that, the ratio of any 2 sides of a triangle is eaqual to the ratio of the lengths of the segments formed on its third side, by the angle bisector of the angle formed by those 2 sides.

So, here, AB/AC = BD/DC

=> 6/5 = 3 / DC

=> DC = 15/6 = 5/2 = 2.5 cm

1.8k Views · View 5 Upvoters · Answer requested by Prafful Sharma

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Vivek Pimpalkar

Answered Jul 25, 2018

AB/Ac=bd/CD,6/5=3/CD, cd=2.5

783 Views · View 3 Upvoters

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Answer 2

Answer:

1:2

Step-by-step explanation:

Given: In ΔABC, ∠B = 90° and ∠A = 60°

The angle bisector of ∠BAC meets BC at D.

In ΔBAD, ∠B = 90° and ∠BAD = 30°

$\tan30^\circ=\dfrac{BD}{BA}$--------------(1)

In ΔABC, ∠B = 90° and ∠BAC = 60°

$\tan60^\circ=\dfrac{BC}{BA}$-------------(2)

Using (1) and (2)

$\dfrac{\tan60^\circ}{\tan30^\circ}=\dfrac{BC}{BD}$

$\dfrac{\sqrt{3}}{1/\sqrt{3}}=\dfrac{BD+DC}{BD}$

$3=\dfrac{BD+DC}{BD}$

$3BD=BD+DC$

$2BD=DC$

$\dfrac{BD}{DC}=\dfrac{1}{2}$

Hence, the ratio of BD and DC is 1:2