Sec 4 A-Math · Trigonometric identities and formulae

Ratios from a given ratio questions with answers

Practice questions on ratios from a given ratio, from the basics up to exam level, each with a full worked solution. Try them on paper first, then open the solution to check your working.

Worked example

Given that $\tan A = -\dfrac{7}{24}$ and $A$ is obtuse, find the exact value of $\cos A$ .

Step 1 of 3

Draw a right-angled triangle with the sizes in tan A (ignore the sign), and find the hypotenuse by Pythagoras' theorem

\text{Hypotenuse} = \sqrt{7^{2} + 24^{2}}

= 25

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Questions

Question 1Acute and obtuse angles
Given that $\sin A = \dfrac{12}{13}$ and $A$ is acute, find the exact value of $\cos A$ .
Show the worked solution
1. Step 1: Draw a right-angled triangle with the sizes in sin A (ignore the sign), and find the adjacent by Pythagoras' theorem $\text{Adjacent} = \sqrt{13^{2} - 12^{2}}$ $= 5$
2. Step 2: Which quadrant is A in? Is cos positive there? (ASTC: All, Sine, Tan, Cos) $A$ is in quadrant 1, so $\cos A$ is positive.
3. Step 3: cos = adjacent ÷ hypotenuse, with that sign $\cos A = \dfrac{5}{13}$
Question 2Acute and obtuse angles
Given that $\tan A = -3$ and $A$ is obtuse, find the exact value of $\cos A$ .
Show the worked solution
1. Step 1: Draw a right-angled triangle with the sizes in tan A (ignore the sign), and find the hypotenuse by Pythagoras' theorem $\text{Hypotenuse} = \sqrt{3^{2} + 1^{2}}$ $= \sqrt{10}$
2. Step 2: Which quadrant is A in? Is cos positive there? (ASTC: All, Sine, Tan, Cos) $A$ is in quadrant 2, so $\cos A$ is negative.
3. Step 3: cos = adjacent ÷ hypotenuse, with that sign (rationalise the denominator) $\cos A = -\dfrac{\sqrt{10}}{10}$
Question 3Reflex angles; sec, cosec and cot
Given that $\tan A = -\dfrac{7}{24}$ and $270^\circ < A < 360^\circ$ , find the exact value of $\cot A$ .
Show the worked solution
1. Step 1: Draw a right-angled triangle with the sizes in tan A (ignore the sign), and find the hypotenuse by Pythagoras' theorem $\text{Hypotenuse} = \sqrt{7^{2} + 24^{2}}$ $= 25$
2. Step 2: Which quadrant is A in? Is cot positive there? (ASTC: All, Sine, Tan, Cos) $A$ is in quadrant 4, so $\cot A$ is negative.
3. Step 3: cot = adjacent ÷ opposite, with that sign $\cot A = -\dfrac{24}{7}$
Question 4Reflex angles; sec, cosec and cot
Given that $\cos A = \dfrac{5}{13}$ and $270^\circ < A < 360^\circ$ , find the exact value of $\sec A$ .
Show the worked solution
1. Step 1: Draw a right-angled triangle with the sizes in cos A (ignore the sign), and find the opposite by Pythagoras' theorem $\text{Opposite} = \sqrt{13^{2} - 5^{2}}$ $= 12$
2. Step 2: Which quadrant is A in? Is sec positive there? (ASTC: All, Sine, Tan, Cos) $A$ is in quadrant 4, so $\sec A$ is positive.
3. Step 3: sec = hypotenuse ÷ adjacent, with that sign $\sec A = \dfrac{13}{5}$

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Worked example

Given that

\tan A = -\dfrac{7}{24}

and

A

is obtuse, find the exact value of

\cos A

Step 1 of 3

Draw a right-angled triangle with the sizes in tan A (ignore the sign), and find the hypotenuse by Pythagoras' theorem

\text{Hypotenuse} = \sqrt{7^{2} + 24^{2}}

= 25

Questions

Question 1Acute and obtuse angles

Given that $\sin A = \dfrac{12}{13}$ and $A$ is acute, find the exact value of $\cos A$ .

Show the worked solution

Step 1: Draw a right-angled triangle with the sizes in sin A (ignore the sign), and find the adjacent by Pythagoras' theorem $\text{Adjacent} = \sqrt{13^{2} - 12^{2}}$ $= 5$
Step 2: Which quadrant is A in? Is cos positive there? (ASTC: All, Sine, Tan, Cos) $A$ is in quadrant 1, so $\cos A$ is positive.
Step 3: cos = adjacent ÷ hypotenuse, with that sign $\cos A = \dfrac{5}{13}$

Question 2Acute and obtuse angles

Given that $\tan A = -3$ and $A$ is obtuse, find the exact value of $\cos A$ .

Show the worked solution

Step 1: Draw a right-angled triangle with the sizes in tan A (ignore the sign), and find the hypotenuse by Pythagoras' theorem $\text{Hypotenuse} = \sqrt{3^{2} + 1^{2}}$ $= \sqrt{10}$
Step 2: Which quadrant is A in? Is cos positive there? (ASTC: All, Sine, Tan, Cos) $A$ is in quadrant 2, so $\cos A$ is negative.
Step 3: cos = adjacent ÷ hypotenuse, with that sign (rationalise the denominator) $\cos A = -\dfrac{\sqrt{10}}{10}$

Question 3Reflex angles; sec, cosec and cot

Given that $\tan A = -\dfrac{7}{24}$ and $270^\circ < A < 360^\circ$ , find the exact value of $\cot A$ .

Show the worked solution

Step 1: Draw a right-angled triangle with the sizes in tan A (ignore the sign), and find the hypotenuse by Pythagoras' theorem $\text{Hypotenuse} = \sqrt{7^{2} + 24^{2}}$ $= 25$
Step 2: Which quadrant is A in? Is cot positive there? (ASTC: All, Sine, Tan, Cos) $A$ is in quadrant 4, so $\cot A$ is negative.
Step 3: cot = adjacent ÷ opposite, with that sign $\cot A = -\dfrac{24}{7}$

Question 4Reflex angles; sec, cosec and cot

Given that $\cos A = \dfrac{5}{13}$ and $270^\circ < A < 360^\circ$ , find the exact value of $\sec A$ .

Show the worked solution

Step 1: Draw a right-angled triangle with the sizes in cos A (ignore the sign), and find the opposite by Pythagoras' theorem $\text{Opposite} = \sqrt{13^{2} - 5^{2}}$ $= 12$
Step 2: Which quadrant is A in? Is sec positive there? (ASTC: All, Sine, Tan, Cos) $A$ is in quadrant 4, so $\sec A$ is positive.
Step 3: sec = hypotenuse ÷ adjacent, with that sign $\sec A = \dfrac{13}{5}$