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
Menu
Grade: 12th pass
        
In triangle ABC ,angle B=90. P is point in a triangle such that angle APB = angle BPC = angle APC. BP=6,AP=10. FInd PC and area of triangle ABC?
2 months ago

Answers : (1)

Aditya Gupta
1669 Points
							
obviously since angle APB = angle BPC = angle APC, they must all be equal to 120 deg.
now, we know that area of a triangle can be expressed as ½ absinC. lets call the area of triangle ABC as A.
then A= sum of areas of triangles APB, BPC, APC
= ½ *6*10*sin120 + ½ *6*x*sin120 + ½ *x*10*sin120, where x= PC.
or A/3= 15+4x........(1)
now, A can also be writen as ½ *AB*BC. 
using cosine rule in triangle ABP, we can easily find AB= 14.
so A= 7*BC.
again using cos rule in triangle BPC, we can write BC in terms of x as:
BC = (x^2+6x+36) so that
A= 7(x^2+6x+36) …..(2)
solving (1) and (2), we get
PC= 33, A= 1473
kindly approve :)
2 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 53 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details