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Please help me with this question

Please help me with this question-example-1
User Gagravarr
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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).

User Dalin
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