Title: Solving Trigonometric Expressions in a Right Triangle
Given the lengths of sides in a right triangle ( \triangle PQR ) where ( PQ = 12 , \text{cm} ), ( PR = 13 , \text{cm} ), and we need to find ( QR ).
Applying Pythagoras Theorem
According to the Pythagorean theorem in a right triangle, the sum of the squares of the two shorter sides is equal to the square of the hypotenuse.
Given:
– ( PQ = 12 , \text{cm} )
– ( PR = 13 , \text{cm} )
Using the Pythagorean theorem:
[ PR^2 = PQ^2 + QR^2 ]
[ 13^2 = 12^2 + QR^2 ]
[ 169 = 144 + QR^2 ]
[ QR^2 = 25 ]
Taking the square root of both sides:
[ QR = 5 , \text{cm} ]
Finding ( \tan{P} – \cot{R} )
To find ( \tan{P} – \cot{R} ), we need to determine the tangent of angle ( P ) and the cotangent of angle ( R ).
Given:
– ( PQ = 12 , \text{cm} )
– ( QR = 5 , \text{cm} )
Using trigonometric ratios:
[ \tan{P} = \frac{PQ}{QR} = \frac{12}{5} ]
[ \cot{R} = \frac{QR}{PQ} = \frac{5}{12} ]
Substitute the values:
[ \tan{P} – \cot{R} = \frac{12}{5} – \frac{5}{12} = \frac{144}{60} – \frac{25}{60} = \frac{119}{60} ]
Therefore, ( \tan{P} – \cot{R} = \frac{119}{60} ) or approximately 1.9833.
Conclusion
In this problem, we used the Pythagorean theorem to find the missing side length ( QR ) in a right triangle. Furthermore, we calculated the difference between the tangent of angle ( P ) and the cotangent of angle ( R ) to obtain the value of ( \tan{P} – \cot{R} = \frac{119}{60} ). This process demonstrates the application of trigonometric concepts in solving geometric problems involving right triangles.
