Home
Class 14
MATHS
In triangle ABC, AB = 12 cm, angle B = 6...

In triangle ABC, AB = 12 cm, `angle B = 60^@`, the perpendicular from A to BC meets it at D. The bisector of `angle ABC` meets AD at E. Then E divides AD in the ratio

A

`3 :1`

B

`6 :1`

C

`1 :1`

D

`2 :1`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, let's break it down: ### Step 1: Understand the Triangle We have triangle ABC with: - AB = 12 cm - Angle B = 60 degrees ### Step 2: Draw the Perpendicular A perpendicular line is drawn from point A to line BC, meeting at point D. This means that angle ADB is a right angle (90 degrees). ### Step 3: Use Cosine to Find AD In triangle ABD, we can use the cosine of angle B to find the length of AD. - Cosine of angle B (60 degrees) is given by: \[ \cos(60^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AD}{AB} \] - Substituting the known values: \[ \cos(60^\circ) = \frac{AD}{12} \] - Since \(\cos(60^\circ) = \frac{1}{2}\), we have: \[ \frac{1}{2} = \frac{AD}{12} \] - Solving for AD: \[ AD = 12 \times \frac{1}{2} = 6 \text{ cm} \] ### Step 4: Apply the Angle Bisector Theorem According to the angle bisector theorem, the ratio in which point E divides AD is given by: \[ \frac{AE}{ED} = \frac{AB}{BD} \] - Here, AB = 12 cm and we need to find BD. ### Step 5: Find BD In triangle ABD, since AD is the height and we already found AD = 6 cm, we can find BD using the sine of angle B: - We know that: \[ \sin(60^\circ) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BD}{AB} \] - Substituting the known values: \[ \sin(60^\circ) = \frac{BD}{12} \] - Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), we have: \[ \frac{\sqrt{3}}{2} = \frac{BD}{12} \] - Solving for BD: \[ BD = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3} \text{ cm} \] ### Step 6: Calculate the Ratio AE/ED Now we can substitute the values into the angle bisector theorem: \[ \frac{AE}{ED} = \frac{AB}{BD} = \frac{12}{6\sqrt{3}} = \frac{2}{\sqrt{3}} \] To express this ratio in a simpler form, we can rationalize the denominator: \[ \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \] ### Step 7: Final Ratio Thus, the ratio AE:ED can be expressed as: \[ AE:ED = 2:3 \] ### Conclusion The final answer is that E divides AD in the ratio of 2:1.
Promotional Banner

Similar Questions

Explore conceptually related problems

In triangle ABC , AB=7 cm, BC = 10 cm, and AC = 8 cm. If AD is the angle bisector of angle BAC , where D is a point on BC, then BD is equal to:

In a triangle ABC, angle B=90^(@), AB=8 cm, BC=6cm . BD is perpendicular to AC, then find the length of CD, AD and BD.

ABC is a right triangle with AB = AC.If bisector of angle A meet BC at D then prove that BC =2 AD .

In triangle ABC , angle C=90^@ ,AC = 5cm and BC = 12 cm. The bisector of angle A meets BC at D. What is the length of AD? triangle ABC में, angle C=90^@ , AC = 5 सेमी तथा BC = 12 सेमी है | कोण A का समद्विभाजक BC से D पर मिलता है | AD की लंबाई कितनी है ?

Assertion (A) : AD is angle bisector of angleA of the triangle ABC. If AB = 6 cm, BC = 7 cm, AC = 8 cm then BD = 3 cm and CD = 4 cm. Reason (R) : The angle bisector AD of the triangle divides base BC in the ratio AB : AC.

AB of Delta ABC is produced to D.CE, the bisector of angle ACB,meets AB at E .show that /_CED=(/_CAB+/_CBD)/(2)

In ABC,/_A=50^(@) and BC is produced to a point D.The bisectors of /_ABC and /_ACD meet at E .find /_E .