Question

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How many distinct triangles ABC is there, up to similarity, such that the magnitudes of angles A, B, and C in degrees are positive integers and satisfy - cos A cos B +sin A sin B sin kC =1, for some positive integer k, where KC does not exceed 360° ?


Please answer in detail and derive the trigonometric formulas used to solve -


Best Of Luck --




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Answer 1

Given:  magnitudes of angles A, B, and C in degrees are positive integers and satisfy  cosA cosB + sinA sinB sinkC = 1

To find : Number of Distinct triangles ABC up to similarity,

Solution:

cosA cosB + sinA sinB sinkC = 1

Adding and subtrating sinA sinB

=> cosA cosB + sinA sinB – sinA sinB + sinA sinB sinkC = 1

Now  cosA cosB + sinA sinB = Cos(A -B)

=>  cos(A – B)   + sinA sinB (SinkC - 1)  = 1

A & B are positive integers    => SinA . Sin B  is + ve

SinkC - 1  ≤ 0     & cos(A – B)   ≤ 1

So only possible case is that

cos(A – B) = 1    &   SinkC - 1  = 0

=> A - B = 0       &   KC = 90°

=> A = B      & KC = 90°

Number of factors of 90º

90  -= 2 * 3 * 3 * 5

90 = 2¹ × 3² × 5¹

No. of factors = (1 + 1) * ( 2+ 1)  * ( 1+ 1)  

= 2 * 3 * 2  = 12  

12 possible values of  C

1  , 2  , 3  , 5  , 6  , 9 , 10  , 15 , 18  , 30 , 45 , 90

but C can not be odd  as  A = B

Hence A  + B   ( even)  = 180 - C     ( so C must be even)

possible values of C

2  , 6  , 10  , 18  , 30 , 90

for each C only one value of A & B exist

Hence Possible Triangles are 6

A      B      C

89    89     2  

87    87     6

85    85    10

81     81     18

75    75     30

45    45     90

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how many distinct Triangles ABC are there two similarity such that ...

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Answer 2

GIVEN :-

cosA.cosB + sinA.sinB.sinkC = 1

TO FIND :-

Thr number of distinct triangles ABC is there, up to similarity, such that the magnitudes of angles A, B, and C in degrees are positive integers and satisfy - cos A cos B +sin A sin B sin kC =1, for some positive integer k, where KC does not exceed 360°.

SOLUTION :-

We can write cosA.cosB + sinA.sinB.sinkC = 1 as :-

cosA.cosB + sinA.sinB + sinA.sinB.sinkC - sinAsinB = 1

⇒ sinA.sinB(sinkC - 1) = 1 - cosA.cosB + sinA.sinB

⇒ sinA.sinB(sinkC - 1) = 1 - cos(A - B)

[As cos(A - B) = cosA.cosB + sinA.sinB]*

⇒cos(A - B) + sinA.sinB(sinkC - 1) = 1

In the LHS,

sinA.sinB is a positive value, as angles are positive. To obtain 1 in the RHS, one of the values must be 0. So, it can be only be justified if cos(A - B) = 1 and (sinkC - 1) = 0.

sinkC - 1 = 0

⇒sinkC = 1

⇒sinkC = sin 90°

⇒kC = 90°

$\rule{59}{2}$

cos(A - B) = 1

⇒cos(A - B) = cos 0°

⇒A - B = 0°

⇒A = B

Number of factors of 90 :-

90 = (2)(3²)(5)

Then number of factors = (2)(3)(2) = 12

All the factors are :-

1 , 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90

☛ We know, odd numbers and even numbers can never cancel each other to 0. (Example, 4 - 3 ≠ 0) Therefore, A and B are even.

☛ Sum of 2 even numbers is always even.

So, A + B = 180° - C          

(Where C has to be even for getting an even number in the RHS)

So, C can be :- 2, 6, 10, 18, 30, 90

For C = 2, 6, 10, 18, 30, 90, A = B

So, 6 such triangles can be formed which meet the required criteria.            $\Large\rm{\boxed{\cdots{\color{gold} {ANSWER}}}}$

$\underline{\rule{200}{2}}$

*Derivation of cos(A - B) :-

We know cos(A + B) = cosAcosB - sinAsinB

Now,

cos(A - B) = cosAcos(-B) - sinAsin(-B)

⇒cos(A - B) = cosAcosB - sinA × (-sinB)

☛ [As cos(-θ) = cosθ and sin(-θ) = -sinθ]

⇒cos(A - B) = cosAcosB + sinAsinB

☛ [As (-) × (-) = (+)]