Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
In a right angled triangel the hypotenuse is four times as long as the perpendicular drawn to it from the opposite vertex .One of the acute angel is
c2 = h2 + (a-x)2
b2 = h2 + x2
on adding, we have 2h2 + x2 + (a-x)2 = a2 ..(1)
also a = 4h and cos B = b/a = x/b
so x = b2/a
on solving (1), we have h2 + x2 -4hx = 0
substituting above gives
b4 + a4/16 -a2.b2 =0 .. (2)
as cos B = b/a
on solving eqn. (2) we have , Cos2B = 0.933 or 0.067
Cos B = 0.966 or 0.259
B = 15 or 75 degrees
as B is acute , B is 15 degrees
--
regards
Ramesh
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !