To find the values of sin P, cos P, and tan P in the right-angled triangle PQR, given that PR + QR = 30 cm and PQ = 10 cm, we will begin by identifying the sides of the triangle and then moving on to calculating the trigonometric ratios.
Here's a step-by-step solution:
1. Identify the sides of the triangle:
- PR is the hypotenuse.
- QR is the side opposite angle P.
- PQ is the side adjacent to angle P.
2. Given conditions:
- The sum of PR and QR is 30 cm.
- PQ (adjacent side) is 10 cm.
3. Set up the equations:
- Let PR = x and QR = y.
- We have x + y = 30 (Equation 1).
4. Right-angled triangle relationships:
- Using the Pythagorean theorem: [tex]\( PQ^2 + QR^2 = PR^2 \)[/tex].
- Substitute [tex]\( PQ = 10 \)[/tex]: [tex]\( 10^2 + QR^2 = PR^2 \)[/tex].
- This simplifies to: [tex]\( 100 + y^2 = x^2 \)[/tex].
5. Solve the system of equations:
- Using equation 1: [tex]\( y = 30 - x \)[/tex].
- Substitute [tex]\( y = 30 - x \)[/tex] into [tex]\( 100 + y^2 = x^2 \)[/tex]:
[tex]\[ 100 + (30 - x)^2 = x^2 \][/tex]
- Solving the algebraic equation results in: [tex]\( x = 16.67 \)[/tex].
6. Determine other side:
- Given [tex]\( PR = x = 16.67 \)[/tex], calculate the length of QR:
[tex]\[ QR = 30 - PR \][/tex]
[tex]\[ QR = 30 - 16.67 \][/tex]
[tex]\[ QR = 13.33 \][/tex]
7. Trigonometric ratios:
- Sin P (Opposite/Hypotenuse):
[tex]\[ \sin P = \frac{QR}{PR} = \frac{13.33}{16.67} \approx 0.7996 \][/tex]
- Cos P (Adjacent/Hypotenuse):
[tex]\[ \cos P = \frac{PQ}{PR} = \frac{10}{16.67} \approx 0.5999 \][/tex]
- Tan P (Opposite/Adjacent):
[tex]\[ \tan P = \frac{QR}{PQ} = \frac{13.33}{10} = 1.333 \][/tex]
To summarize, the values of the trigonometric ratios for angle P in the triangle PQR are as follows:
- [tex]\( \sin P \approx 0.7996 \)[/tex]
- [tex]\( \cos P \approx 0.5999 \)[/tex]
- [tex]\( \tan P = 1.333 \)[/tex]
