S L Loney Solutions for Chapter: Relations Between the Sides and the Trigonometrical Ratios of the Angles of a Triangle, Exercise 2: Examples XXVII

Author:S L Loney

S L Loney Mathematics Solutions for Exercise - S L Loney Solutions for Chapter: Relations Between the Sides and the Trigonometrical Ratios of the Angles of a Triangle, Exercise 2: Examples XXVII

Attempt the practice questions on Chapter 11: Relations Between the Sides and the Trigonometrical Ratios of the Angles of a Triangle, Exercise 2: Examples XXVII with hints and solutions to strengthen your understanding. Plane Trigonometry Part 1 solutions are prepared by Experienced Embibe Experts.

Questions from S L Loney Solutions for Chapter: Relations Between the Sides and the Trigonometrical Ratios of the Angles of a Triangle, Exercise 2: Examples XXVII with Hints & Solutions

HARD
JEE Main
IMPORTANT

The sides of a triangle are in A.P. and the greatest and the least angles are θ and ϕ, prove that 41-cosθ1-cosϕ=cosθ+cosϕ.

HARD
JEE Main
IMPORTANT

 The sides of a triangle are in A.P. and the greatest angle exceeds the least by 90°; prove that the sides are proportional to 7+1, 7 and 7-1.

MEDIUM
JEE Main
IMPORTANT

If in a ABC, C=60°, then prove that 1a+c+1b+c=3a+b+c.

HARD
JEE Main
IMPORTANT

In any triangle ABC, if D be any point of the base BC, such that BD:DC::m:n and if BAD=α, DAC=β, CDA=θ and AD=x, prove that m+ncotθ=mcotα-ncotβ=ncotB-mcotC and m+n2:x2=m+nmb2+nc2-mna2.

MEDIUM
JEE Main
IMPORTANT

If in a triangle the bisector of the side c be perpendicular to the side b, prove that 2tanA+tanC=0.

EASY
JEE Main
IMPORTANT

In any triangle prove that if θ be any angle, then bcosθ=ccosA-θ+acosC+θ.

MEDIUM
JEE Main
IMPORTANT

If p and q be the perpendiculars from the angular points A and B on any line passing through the vertex C of the triangle ABC, then prove that
a2p2+b2q2-2abpqcosC=a2b2sin2C.

HARD
JEE Main
IMPORTANT

In the triangle ABC, lines OA, OB and OC are drawn so that the angle OAB, OBC and OCA are each equal to ω; prove that cotω=cotA+cotB+cotC and cosec2ω=cosec2A+cosec2B+cosec2C.