From the diagram, Find the value of 1) Sin x, Cos x and tan x ii) Sin x - Cos x iii) (Sin x + tan x)", Solve

We are given the side lengths $PQ = 8$ cm, $PR = 6$ cm, and $QR = 10$ cm. We check if it's a right-angled triangle using the Pythagorean theorem ($a^2 + b^2 = c^2$).

$$PQ^2 + PR^2 = 8^2 + 6^2 = 64 + 36 = 100$$ $$QR^2 = 10^2 = 100$$

Since $PQ^2 + PR^2 = QR^2$, the triangle PQR is a right-angled triangle with the right angle at P.

For angle $x$ at vertex Q:

• The side opposite to $x$ is $PR = 6$ cm.
• The side adjacent to $x$ is $PQ = 8$ cm.
• The hypotenuse is $QR = 10$ cm.

Using SOH CAH TOA:

$$\sin x = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{PR}{QR} = \frac{6}{10} = \frac{3}{5}$$ $$\cos x = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{PQ}{QR} = \frac{8}{10} = \frac{4}{5}$$ $$\tan x = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{PR}{PQ} = \frac{6}{8} = \frac{3}{4}$$

So, for part i):
$\boxed{\sin x = \frac{3}{5}, \cos x = \frac{4}{5}, \tan x = \frac{3}{4}}$

Substitute the values found in Step 2:

$$\sin x - \cos x = \frac{3}{5} - \frac{4}{5} = \frac{3-4}{5} = -\frac{1}{5}$$

So, for part ii):
$\boxed{\sin x - \cos x = -\frac{1}{5}}$

Substitute the values found in Step 2:

$$\sin x + \tan x = \frac{3}{5} + \frac{3}{4}$$

Find a common denominator, which is 20:

$$\frac{3}{5} + \frac{3}{4} = \frac{3 \times 4}{5 \times 4} + \frac{3 \times 5}{4 \times 5} = \frac{12}{20} + \frac{15}{20} = \frac{12+15}{20} = \frac{27}{20}$$

Now, square the result:

$$\left(\frac{27}{20}\right)^2 = \frac{27^2}{20^2} = \frac{729}{400}$$

So, for part iii):
$\boxed{(\sin x + \tan x)^2 = \frac{729}{400}}$