Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

In a triangle ABC, angle A is twice the angle B. show that a 2 =b(b+c),where a,b,and c are the sides opposite to angle A,B and,C respectively .

In a triangle ABC, angle A is twice the angle B. show that a2 =b(b+c),where a,b,and c are the sides opposite to angle A,B and,C respectively.

Grade:10

1 Answers

Saurabh Koranglekar
askIITians Faculty 10341 Points
one year ago
Dear Student

Given, angle A = 2*angle B,
Apply sine rule i.e..

a/(Sin A)= b/(Sin B)= c/(Sin c)

Take the first 2 entities as we were given information about A and B. Also, substitute A = 2B
Now, we have
a/(Sin 2B) = b/(Sin B)
Sin 2B = 2 *(Sin B)*(Cos B)

Substituting this in the above equation. Thus,
Sin B gets canceled because B is not equal to zero(because it is a triangle). By extension. Sin B is not equal to zero.

We get,
a/(2 Cos B) = b [now apply cosine rule]
Cos B = (a^2 + c^2 - b^2)/2*a*c
Substituting we get the result,
a^2 = b(b+c)=b^2+b*c
BC^2 = AC^2 + AC*AB

Regards

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free