Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Offsite Campus Turito Championship

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Maths-

General

Easy

Question

If the sides be a, b, c then (r + r1) tan $fraction numerator B minus C over denominator 2 end fraction$ +(r + r2) tan $fraction numerator C minus A over denominator 2 end fraction$ + (r+ r3) tan $fraction numerator A minus B over denominator 2 end fraction$ =

1
0
2
4

The correct answer is: 0

Related Questions to study

$open parentheses fraction numerator 1 over denominator r subscript 1 end subscript end fraction plus fraction numerator 1 over denominator r subscript 2 end subscript end fraction close parentheses$ $open parentheses fraction numerator 1 over denominator r subscript 3 end subscript end fraction plus fraction numerator 1 over denominator r subscript 1 end subscript end fraction close parentheses$ =

$open parentheses fraction numerator 1 over denominator r subscript 1 end subscript end fraction plus fraction numerator 1 over denominator r subscript 2 end subscript end fraction close parentheses$ $open parentheses fraction numerator 1 over denominator r subscript 3 end subscript end fraction plus fraction numerator 1 over denominator r subscript 1 end subscript end fraction close parentheses$ =

$I f space fraction numerator left parenthesis x minus 2 right parenthesis left parenthesis 1 minus x right parenthesis left parenthesis x minus 3 right parenthesis cubed left parenthesis x minus 4 right parenthesis squared over denominator left parenthesis x plus 1 right parenthesis end fraction less or equal than 0 space t h e n space x space l i e s space i n$

$I f space fraction numerator left parenthesis x minus 2 right parenthesis left parenthesis 1 minus x right parenthesis left parenthesis x minus 3 right parenthesis cubed left parenthesis x minus 4 right parenthesis squared over denominator left parenthesis x plus 1 right parenthesis end fraction less or equal than 0 space t h e n space x space l i e s space i n$

(r1 – r) (r2 – r) (r3 – r) =

(r1 – r) (r2 – r) (r3 – r) =

parallel

$fraction numerator 1 over denominator r subscript 1 end subscript superscript 2 end superscript end fraction$ + $fraction numerator 1 over denominator r subscript 2 end subscript superscript 2 end superscript end fraction$ + $fraction numerator 1 over denominator r subscript 3 end subscript superscript 2 end superscript end fraction$ + $fraction numerator 1 over denominator r subscript blank superscript 2 end superscript end fraction$ =

$fraction numerator 1 over denominator r subscript 1 end subscript superscript 2 end superscript end fraction$ + $fraction numerator 1 over denominator r subscript 2 end subscript superscript 2 end superscript end fraction$ + $fraction numerator 1 over denominator r subscript 3 end subscript superscript 2 end superscript end fraction$ + $fraction numerator 1 over denominator r subscript blank superscript 2 end superscript end fraction$ =

$I f space fraction numerator left parenthesis 2 x minus 1 right parenthesis left parenthesis x minus 1 right parenthesis squared left parenthesis x minus 2 right parenthesis cubed over denominator left parenthesis x minus 4 right parenthesis to the power of 4 end fraction greater than 0 space t h e n space x space l i e s space i n$

$I f space fraction numerator left parenthesis 2 x minus 1 right parenthesis left parenthesis x minus 1 right parenthesis squared left parenthesis x minus 2 right parenthesis cubed over denominator left parenthesis x minus 4 right parenthesis to the power of 4 end fraction greater than 0 space t h e n space x space l i e s space i n$

$i f space fraction numerator open parentheses x squared plus 3 close parentheses open parentheses x minus 1 close parentheses to the power of 4 over denominator open parentheses x plus 1 close parentheses end fraction greater than 0 space t h e n space x element of$

$i f space fraction numerator open parentheses x squared plus 3 close parentheses open parentheses x minus 1 close parentheses to the power of 4 over denominator open parentheses x plus 1 close parentheses end fraction greater than 0 space t h e n space x element of$

parallel

If (a – b) (s – c) = (b – c) (s – a) then r1, r2, r3 are in -

If (a – b) (s – c) = (b – c) (s – a) then r1, r2, r3 are in -

If in a triangle $open parentheses 1 minus fraction numerator r subscript 1 end subscript over denominator r subscript 3 end subscript end fraction close parentheses$ = 2, then the triangle is-

If in a triangle $open parentheses 1 minus fraction numerator r subscript 1 end subscript over denominator r subscript 3 end subscript end fraction close parentheses$ = 2, then the triangle is-

The angles of a right-angled triangle are in A.P. The ratio of the in-radius and the perimeter is -

The angles of a right-angled triangle are in A.P. The ratio of the in-radius and the perimeter is -

parallel

In a ABC, if $fraction numerator R over denominator r end fraction$  2, then the triangle is -

In a ABC, if $fraction numerator R over denominator r end fraction$  2, then the triangle is -

abc =

abc =

If the altitudes of a triangle be 3, 4, 6 then its in-radius is -

If the altitudes of a triangle be 3, 4, 6 then its in-radius is -

parallel

4Rr cos $fraction numerator A over denominator 2 end fraction$ cos $fraction numerator B over denominator 2 end fraction$ cos $fraction numerator C over denominator 2 end fraction$ =

4Rr cos $fraction numerator A over denominator 2 end fraction$ cos $fraction numerator B over denominator 2 end fraction$ cos $fraction numerator C over denominator 2 end fraction$ =

If the area of a triangle is 81 square cm and its perimeter is 27cm then its in-radius in centi-metres is -

If the area of a triangle is 81 square cm and its perimeter is 27cm then its in-radius in centi-metres is -

If in a triangle ABC, C = 90° then the maximum value of sin A sin B is -

If in a triangle ABC, C = 90° then the maximum value of sin A sin B is -

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.