Question

Please help me with this question

Answer 1

In the first triangle with sides 3 cm, 4 cm, and 5 cm, trigonometric ratios for angles A and B are
`\(\sin(A) = (3)/(5)\), \(\cos(A) = (4)/(5)\), \(\tan(A) = (3)/(4)\)`. In the second triangle, ratios are
`\(\sin(A) = (10)/(13)\), \(\cos(A) = (√(69))/(13)\), \(\tan(A) = (10)/(√(69))\).`

Let's use the Pythagorean Theorem to find the missing side in each right triangle and then determine the trigonometric ratios for angles A and B.

### Part a:

Given right triangle ACB with BC = 3 cm and AC = 4 cm.

Using the Pythagorean Theorem:
`\(AB^2 = AC^2 + BC^2\)`

`\[AB^2 = 4^2 + 3^2\]`

`\[AB^2 = 16 + 9\]`

`\[AB^2 = 25\]`

AB = 5

Now, we have all three sides of the triangle: AC = 4 cm, BC = 3 cm, and AB = 5 cm.

The trigonometric ratios for angle A and B are as follows:

- For angle A:

-
`\(\sin(A) = (BC)/(AB) = (3)/(5)\)`

-
`\(\cos(A) = (AC)/(AB) = (4)/(5)\)`

-
`\(\tan(A) = (BC)/(AC) = (3)/(4)\)`

- For angle B (since it's a right triangle,
`\(\sin(B)\), \(\cos(B)\), and \(\tan(B)\)` are the same as for angle A).

### Part b:

Given right triangle ACB with BC = 10 cm and AB = 13 cm.

Using the Pythagorean Theorem:
`\(AC^2 = AB^2 - BC^2\)`

`\[AC^2 = 13^2 - 10^2\]`

`\[AC^2 = 169 - 100\]`

`\[AC^2 = 69\]`

`\[AC = √(69)\]`

Now, we have all three sides of the triangle: AC =
`\(√(69)\)` cm, BC = 10 cm, and AB = 13 cm.

The trigonometric ratios for angle A and B are as follows:

- For angle A:

-
`\(\sin(A) = (BC)/(AB) = (10)/(13)\)`

-
`\(\cos(A) = (AC)/(AB) = (√(69))/(13)\)`

-
`\(\tan(A) = (BC)/(AC) = (10)/(√(69))\)`

- For angle B (since it's a right triangle,
`\(\sin(B)\), \(\cos(B)\), and \(\tan(B)\)` are the same as for angle A).