In , right-angled at Q, and . Determine the values of , and .

Understand the Question In right with , is the hypotenuse, while and are the legs. Given and . Letting , we have . Using the Pythagoras theorem (), we find the lengths of all three sides. Finally, we calculate the trigonometric ratios with respect to : Step 1 · Find Side Lengths of Consider right triangle with . Given and . Let , then . By Pythagoras theorem in Therefore Step 2 · Calculate Trigonometric Ratios For , the opposite side is , the adjacent side is , and the hypotenuse is . Answer > , , Common Mistakes Algebraic Expansion Error: Incorrectly expanding as instead of . Side Identification: Swapping opposite and adjacent sides for . Remember that side is opposite to and is adjacent to .

Introduction to Trigonometry | Exercise 8.1

Question 10

In $\Delta \text{PQR}$ , right-angled at Q, $\text{PR} + \text{QR} = 25\text{ cm}$ and $\text{PQ} = 5\text{ cm}$ . Determine the values of $\sin P$ , $\cos P$ and $\tan P$ .

Solution

Understand the Question

In right $\Delta \text{PQR}$ with $\angle \text{Q} = 90^\circ$ , $\text{PR}$ is the hypotenuse, while $\text{PQ}$ and $\text{QR}$ are the legs.
Given $\text{PQ} = 5\text{ cm}$ and $\text{PR} + \text{QR} = 25\text{ cm}$ . Letting $\text{QR} = x$ , we have $\text{PR} = 25 - x$ .
Using the Pythagoras theorem ( $\text{PR}^2 = \text{PQ}^2 + \text{QR}^2$ ), we find the lengths of all three sides.
Finally, we calculate the trigonometric ratios with respect to $\angle P$ $∠ P$ :
- $\sin P = \dfrac{\text{Opposite}}{\text{Hypotenuse}} = \dfrac{\text{QR}}{\text{PR}}$
- $\cos P = \dfrac{\text{Adjacent}}{\text{Hypotenuse}} = \dfrac{\text{PQ}}{\text{PR}}$
- $\tan P = \dfrac{\text{Opposite}}{\text{Adjacent}} = \dfrac{\text{QR}}{\text{PQ}}$

Step 1 · Find Side Lengths of $\Delta \text{PQR}$

Consider right triangle $\text{PQR}$ with $\angle \text{Q} = 90^\circ$ . Diagram 1

Given $\text{PQ} = 5\text{ cm}$ and $\text{PR} + \text{QR} = 25\text{ cm}$ .

Let $\text{QR} = x$ , then $\text{PR} = 25 - x$ .

By Pythagoras theorem in $\Delta \text{PQR}$

\begin{aligned} \text{PR}^2 &= \text{PQ}^2 + \text{QR}^2 \\ (25 - x)^2 &= (5)^2 + (x)^2 \\ 625 - 50x + x^2 &= 25 + x^2 \\ 625 - 50x &= 25 \\ 625 - 25 &= 50x \\ 600 &= 50x \\ x &= \dfrac{600}{50} \\[0.6em] x &= 12 \end{aligned}

Therefore $\text{QR} = 12\text{ cm}$

$\text{PR} = 25 - 12 = 13\text{ cm}$

Step 2 · Calculate Trigonometric Ratios

For $\angle P$ , the opposite side is $\text{QR} = 12\text{ cm}$ , the adjacent side is $\text{PQ} = 5\text{ cm}$ , and the hypotenuse is $\text{PR} = 13\text{ cm}$ .

\begin{aligned} \sin P &= \dfrac{\text{Opposite}}{\text{Hypotenuse}} = \dfrac{\text{QR}}{\text{PR}} = \dfrac{12}{13} \end{aligned}

\begin{aligned} \cos P &= \dfrac{\text{Adjacent}}{\text{Hypotenuse}} = \dfrac{\text{PQ}}{\text{PR}} = \dfrac{5}{13} \end{aligned}

\begin{aligned} \tan P &= \dfrac{\text{Opposite}}{\text{Adjacent}} = \dfrac{\text{QR}}{\text{PQ}} = \dfrac{12}{5} \end{aligned}

Answer

$\sin P = \dfrac{12}{13}$ , $\cos P = \dfrac{5}{13}$ , $\tan P = \dfrac{12}{5}$

Common Mistakes

Algebraic Expansion Error: Incorrectly expanding $(25 - x)^2$ as $625 - x^2$ instead of $625 - 50x + x^2$ .
Side Identification: Swapping opposite and adjacent sides for $\angle P$ . Remember that side $\text{QR}$ is opposite to $\angle P$ and $\text{PQ}$ is adjacent to $\angle P$ .

Question 10

More questions in Exercise 8.1