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# Geometry and Trigonometry | PRMO 2019 | Problem 11

Try this beautiful problem from Pre RMO, 2019 based on Geometry and Trigonometry.

## Geometry and Trigonometry - PRMO 2019

How many distinct triangles ABC are there, up to similarity, such that the magnitudes of angle A, B and C in degrees are positive integers and satisfy cosAcosB + sinAsinBsinkC=1 for some positive integer k, where kC does not exceed 360 degrees.

• is 13
• is 25
• is 6
• cannot be determined from the given information

### Key Concepts

Geometry

Trigonometry

Number Theory

PRMO, 2019

Plane Trigonometry by Loney

## Try with Hints

First hint

Here cosAcosB+sinAsinBsinkC=1 then cosAcosB+sinAsinB+sinAsinBsinkC-sinAsinB=1 then sinAsinB(sinkC-1)=1-cos(A-B)

Second Hint

Then sinkC-1=0 and cos(A-B)=1 then kC=90 and A=B

Final Step

Then Number of factors of 90 is 90=(2)($3^{2}$)(5) then number of factors=(2)(3)(2)=12 for 6 factor A,B are integers

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Try this beautiful problem from Pre RMO, 2019 based on Geometry and Trigonometry.

## Geometry and Trigonometry - PRMO 2019

How many distinct triangles ABC are there, up to similarity, such that the magnitudes of angle A, B and C in degrees are positive integers and satisfy cosAcosB + sinAsinBsinkC=1 for some positive integer k, where kC does not exceed 360 degrees.

• is 13
• is 25
• is 6
• cannot be determined from the given information

### Key Concepts

Geometry

Trigonometry

Number Theory

PRMO, 2019

Plane Trigonometry by Loney

## Try with Hints

First hint

Here cosAcosB+sinAsinBsinkC=1 then cosAcosB+sinAsinB+sinAsinBsinkC-sinAsinB=1 then sinAsinB(sinkC-1)=1-cos(A-B)

Second Hint

Then sinkC-1=0 and cos(A-B)=1 then kC=90 and A=B

Final Step

Then Number of factors of 90 is 90=(2)($3^{2}$)(5) then number of factors=(2)(3)(2)=12 for 6 factor A,B are integers

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