Question

in a right angled triangle ABC, sin(A+B)=1, cos A=1/2, find
cot(C-B).​

Answer 1

Answer:

Angles of ‘A’, ‘B’ and ‘C’ are \bold{67.5^{\circ}, 37.5^{\circ}, 75^{\circ}}67.5

∘

,37.5

∘

,75

∘

GIVEN:

A+B+C = 1/2

To find:

Angles of A,B nd C

Solution:

Let us take the triangle ABC, now as the question says A+B-C=\frac{1}{2}\ \text{and}\ \text{cos} B+C-A=\frac{1}{\sqrt{2}}A+B−C=

2

1

and cosB+C−A=

2

1

Therefore, the values can be written, as

\sin A+B-C=\frac{1}{2} ; \quad A+B-C=\sin ^{-1}\left(\frac{1}{2}\right) ; A+B-C=30sinA+B−C=

2

1

;A+B−C=sin

−1

(

2

1

);A+B−C=30 and for the cosine the value is B+C-A=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right) ; B+C-A=45B+C−A=cos

−1

(

2

1

);B+C−A=45

So, as we know that the sum of all ‘sides of a triangle’ is 180,

Hence, A + B + C = 180.A+B+C=180.

With all the equation, we can find the values of A, B and C. Equating all equation:

Let us take A+B-C=30;A+B+C=180A+B−C=30;A+B+C=180

Substituting value of A+B=C+30 in A+B+C=180, we get C+30+C=180;C=75^{\circ}C+30+C=180;C=75

∘

Now with C=75^{\circ}C=75

∘

we can find the value of A and B, putting the value of C in A+B=C+30 we get A+B=75+30; A+B=105A+B=75+30;A+B=105

And putting the value of C\ \text{in}\ B+C-A=45C in B+C−A=45 we get B+75-A=45; B-A=-30 or A-B=30

Equate A-B=30 & A+B=105 we get the value of A as 67.5^{\circ}67.5

∘

Therefore, value of B is 37.5^{\circ}37.5

∘

Hence, the angles of ‘A’, ‘B’ and ‘C’ are 67.5^{\circ}, 37.5^{\circ}, 75^{\circ}67.5

∘

,37.5

∘

,75

∘