if ` vec a ,\ vec b ,\ vec c`
are three mutually perpendicular vectors of equal
 magniltgude, prove that ` vec a+ vec b+ vec c`
is equally inclined with vectors ` vec a ,\ vec b ,\ a n d\ vec cdot`
also find the angle.
solution: `abs(veca)=abs(vecb)=abs(vecc)=lamda`   

`veca*vecb=vecb*vecc*vecc*veca=0`   

`abs(veca+vecb+vecc)^2=(veca+vecb+vecc)*(veca+vecb+vecc)`   

`=abs(veca)^2+0+0+abs(vecb)^2+abs(vecc)^2`   

`=lambda^2+lambda^2+lambda^2`   

`=3lambda^2`   

`abs(veca+vecb+vecc)=sqrt3lambda`   

`costheta\_1=(veca*(veca+vecb+vecc))/(abs(veca)abs(veca+vecb+vecc))=lambda^2/(lambdasqrt3lambda)=1/sqrt3=theta\_1=cos^(-1)(1/sqrt3)`   

similarly,   

`theta\_1=theta\_2=theta\_3=cos^(-1)(1/sqrt3)`
let a and b be the two given vectorsɵ = 150° is the angle between a and bif c is the resultant vector, then: ...1also, it is given that the resultant is perpendicular to one of the given vectors. let that given vector be anow, …2putting 2 in 1, we get:now, putting c = 10, ɵ = 150° ()so, hence, smaller vector is a = 20 units let abc be the trianglewhere ab=(a-1)2 cmbc=2√a cmca=(a+1) cmab2=(a-1)2=a2+1-2abc2=(2√a)2=4a2ca2=(a+1)2=a2+1+2ahence ab2+bc2=ac2so Δabc is a right angles triangle at b let a, b, c be the sides of any triangle abc. then by applying the sine rule, we get⇒c = k sin csimilarly, b = k sin band a = k sin ahere we will consider rhs, so we getrhs = (b2 – c2) sin asubstituting corresponding values in the above equation, we get⇒ = [( k sin b)2 - ( k sin c)2] sin a⇒ = k2(sin2 b – sin2 c )sin a……….(ii)but,substituting the above values in equation (ii), we get⇒ = k2(sin(b + c) sin(b - c)) sin abut a + b + c = π ⇒ b + c = π –a, so the above equation becomes,⇒ = k2(sin(π –a) sin(b - c))sin abut sin (π - θ) = sin θ⇒ = k2(sin(a) sin(b - c))sin arearranging the above equation we get⇒ = (k sin(a))( sin(b - c))(k sin a)from sine rule, a = k sin a, so the above equation becomes,⇒ = a2 sin(b - c) = rhshence proved given: ab=acto prove: ad>acproof:in ∆abc,∠acd = ∠b + ∠bac= ∠acb + ∠bac …as ∠c = ∠b as ab = ac= ∠cad + ∠cda +∠bac …as ∠acb = ∠cad + ∠cda∴∠acd > ∠cdaso the side opposite to ∠acd is the longest∴ad > ac given : let a , b and c be three students whose chances of solving a problem is given i.e , p(a) = , p(b) = and p(c) = . p() = , p() = and p()= to find: the probability that the problem is solved .here , p(the problem is solved) = 1 - p(the problem is not solved)= 1- p( )= 1 – [p( ) p( ) p( )]= 1- []= 1 - = therefore , the probability that the problem is solved is . given that,l ‖ m and n cuts themlet,∠1 = 65o∠2 = x∠3 = 40o∠1 = ∠4 = 65o (alternate angle) (i)∠3 + ∠4 + ∠5 = 180o (angle sum property)40o + 65o + ∠5 = 180o∠5 = 75onow,∠2 + ∠5 = 180o (linear pair)x + 75o = 180ox = 105o consider the parallelogram abcd with diagonals ac and bd as shown and they intersect at right angles at oa parallelogram is a square if its adjacent angles are 90° and adjacent sides are equalso we have to prove that adjacent sides of given parallelogram are equal and adjacent angles are 90° to prove given parallelogram is a squareconsider δbad and δcdaac = bd … givenab = dc … opposite sides of a parallelogramad is the common sidetherefore, δbad ≅ δcda … sss test for congruency⇒ ∠bad = ∠cda …corresponding angles of congruent triangles …(i)as it is given that abcd is parallelogram⇒ ∠bad + ∠cda = 180° … sum of adjacent angles of a parallelogram is 180°using equation (i)⇒ ∠cda + ∠cda = 180°⇒ 2 × ∠cda = 180°⇒ ∠cda = 90°⇒ ∠bad = 90°thus, adjacent angles are 90° … (iii)consider δaob and δaod∠aob = ∠aod … both 90° because given that diagonals intersect at right anglesod = ob … diagonals of a parallelogram bisect each otherao is the common sidetherefore, δaob ≅ δaod … sas test for congruency⇒ ab = ad … corresponding sides of congruent trianglesthus, adjacent sides are equal … (iv)from statements (iii) and (iv) we can conclude that parallelogram abcd is a squaretherefore, if in a parallelogram, the diagonals are equal in lengths and intersect at right angles, the parallelogram will be a square. a, b and c The given case is SAS (two sides and included angle) so use the Law of Cosines:
  

\(\displaystyle{b}^{{2}}={a}^{{2}}+{c}^{{2}}-{2}{a}{c}{\cos{{B}}}\)
  

\(\displaystyle{b}^{{2}}={5}^{{2}}+{3}^{{2}}-{2}{\left({5}\right)}{\left({3}\right)}{\cos{{92}}}°\)
  

\(\displaystyle{b}^{{2}}={34}-{30}{\cos{{92}}}°\)
  

\(\displaystyle{b}=\sqrt{{{34}-{30}{\cos{{92}}}°}}\)
  

b ~ 5.92 in the figure below, the angle  is shown in the standard position where the point  is on the terminal side of , and  is the distance from the origin to . 

if  is the distance from the origin to  then the trigonometric function of  are
  
  
     
therefore, the trigonometric functions are , , and .