Exercise 8.1 Practice
Trigonometric Ratios & Pythagoras Theorem
Q1: Finding Ratios from Sides
In $\Delta ABC$, right-angled at B, $AB = 5$ cm, $BC = 12$ cm. Determine:
(i) $\sin A, \cos A$
(ii) $\sin C, \cos C$
(i) $\sin A, \cos A$
(ii) $\sin C, \cos C$
Step 1: Find Hypotenuse (AC)
Using Pythagoras theorem: $AC^2 = AB^2 + BC^2$
$AC^2 = 5^2 + 12^2 = 25 + 144 = 169$
$AC = \sqrt{169} = 13$ cm
Using Pythagoras theorem: $AC^2 = AB^2 + BC^2$
$AC^2 = 5^2 + 12^2 = 25 + 144 = 169$
$AC = \sqrt{169} = 13$ cm
Step 2: For Angle A (Opposite=BC=12, Adjacent=AB=5)
$\sin A = \frac{\text{Opp}}{\text{Hyp}} = \frac{12}{13}$
$\cos A = \frac{\text{Adj}}{\text{Hyp}} = \frac{5}{13}$
$\sin A = \frac{\text{Opp}}{\text{Hyp}} = \frac{12}{13}$
$\cos A = \frac{\text{Adj}}{\text{Hyp}} = \frac{5}{13}$
Step 3: For Angle C (Opposite=AB=5, Adjacent=BC=12)
$\sin C = \frac{\text{Opp}}{\text{Hyp}} = \frac{5}{13}$
$\cos C = \frac{\text{Adj}}{\text{Hyp}} = \frac{12}{13}$
$\sin C = \frac{\text{Opp}}{\text{Hyp}} = \frac{5}{13}$
$\cos C = \frac{\text{Adj}}{\text{Hyp}} = \frac{12}{13}$
sin A = 12/13, cos A = 5/13; sin C = 5/13, cos C = 12/13
Q2: Evaluate Expression
In the figure below, find $\tan P - \cot R$. (Given PQ = 8 cm, PR = 17 cm)
Step 1: Find QR
$QR = \sqrt{PR^2 - PQ^2} = \sqrt{17^2 - 8^2}$
$QR = \sqrt{289 - 64} = \sqrt{225} = 15$ cm
$QR = \sqrt{PR^2 - PQ^2} = \sqrt{17^2 - 8^2}$
$QR = \sqrt{289 - 64} = \sqrt{225} = 15$ cm
Step 2: Find Ratios
$\tan P = \frac{\text{Opp}}{\text{Adj}} = \frac{QR}{PQ} = \frac{15}{8}$
$\cot R = \frac{\text{Adj}}{\text{Opp}} = \frac{QR}{PQ} = \frac{15}{8}$
$\tan P = \frac{\text{Opp}}{\text{Adj}} = \frac{QR}{PQ} = \frac{15}{8}$
$\cot R = \frac{\text{Adj}}{\text{Opp}} = \frac{QR}{PQ} = \frac{15}{8}$
Step 3: Evaluate
$\tan P - \cot R = \frac{15}{8} - \frac{15}{8} = 0$
$\tan P - \cot R = \frac{15}{8} - \frac{15}{8} = 0$
0
Q3: Ratio Conversion
If $\sin A = \frac{5}{13}$, calculate $\cos A$ and $\tan A$.
Given $\sin A = \frac{5}{13} = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$.
Let $P = 5k$, $H = 13k$.
Let $P = 5k$, $H = 13k$.
Find Base (B):
$B = \sqrt{H^2 - P^2} = \sqrt{(13k)^2 - (5k)^2}$
$B = \sqrt{169k^2 - 25k^2} = \sqrt{144k^2} = 12k$
$B = \sqrt{H^2 - P^2} = \sqrt{(13k)^2 - (5k)^2}$
$B = \sqrt{169k^2 - 25k^2} = \sqrt{144k^2} = 12k$
$\cos A = \frac{B}{H} = \frac{12k}{13k} = \frac{12}{13}$
$\tan A = \frac{P}{B} = \frac{5k}{12k} = \frac{5}{12}$
$\tan A = \frac{P}{B} = \frac{5k}{12k} = \frac{5}{12}$
cos A = 12/13, tan A = 5/12
Q4: Algebraic Ratio
Given $8 \cot A = 15$, find $\sin A$ and $\sec A$.
$\cot A = \frac{15}{8} = \frac{\text{Base}}{\text{Perpendicular}}$
Let $B=15k, P=8k$.
Let $B=15k, P=8k$.
Find Hypotenuse (H):
$H = \sqrt{P^2 + B^2} = \sqrt{(8k)^2 + (15k)^2}$
$H = \sqrt{64k^2 + 225k^2} = \sqrt{289k^2} = 17k$
$H = \sqrt{P^2 + B^2} = \sqrt{(8k)^2 + (15k)^2}$
$H = \sqrt{64k^2 + 225k^2} = \sqrt{289k^2} = 17k$
$\sin A = \frac{P}{H} = \frac{8k}{17k} = \frac{8}{17}$
$\sec A = \frac{H}{B} = \frac{17k}{15k} = \frac{17}{15}$
$\sec A = \frac{H}{B} = \frac{17k}{15k} = \frac{17}{15}$
sin A = 8/17, sec A = 17/15
Q5: All Ratios
Given $\sec \theta = \frac{5}{3}$, calculate all other trigonometric ratios.
$\sec \theta = \frac{H}{B} = \frac{5}{3}$. Let $H=5k, B=3k$.
$P = \sqrt{H^2 - B^2} = \sqrt{25k^2 - 9k^2} = \sqrt{16k^2} = 4k$.
$\sin \theta = \frac{P}{H} = \frac{4}{5}$
$\cos \theta = \frac{B}{H} = \frac{3}{5}$
$\tan \theta = \frac{P}{B} = \frac{4}{3}$
$\cot \theta = \frac{B}{P} = \frac{3}{4}$
$\text{cosec } \theta = \frac{H}{P} = \frac{5}{4}$
$\cos \theta = \frac{B}{H} = \frac{3}{5}$
$\tan \theta = \frac{P}{B} = \frac{4}{3}$
$\cot \theta = \frac{B}{P} = \frac{3}{4}$
$\text{cosec } \theta = \frac{H}{P} = \frac{5}{4}$
Q6: Proof
If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$, then show that $\angle A = \angle B$.
Consider a triangle $ABC$ with $\angle C = 90^\circ$.
$\cos A = \frac{AC}{AB}$ and $\cos B = \frac{BC}{AB}$.
$\cos A = \frac{AC}{AB}$ and $\cos B = \frac{BC}{AB}$.
Given $\cos A = \cos B \Rightarrow \frac{AC}{AB} = \frac{BC}{AB}$.
$AC = BC$.
In $\Delta ABC$, since sides $AC = BC$, the angles opposite to them must be equal.
$\Rightarrow \angle B = \angle A$.
$\Rightarrow \angle B = \angle A$.
Hence Proved.
Q7: Evaluation
If $\cot \theta = \frac{3}{4}$, evaluate:
(i) $\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$
(ii) $\cot^2 \theta$
(i) $\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$
(ii) $\cot^2 \theta$
Part (i): Numerator = $1 - \sin^2 \theta = \cos^2 \theta$.
Denominator = $1 - \cos^2 \theta = \sin^2 \theta$.
Denominator = $1 - \cos^2 \theta = \sin^2 \theta$.
Expression becomes $\frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta$.
Given $\cot \theta = \frac{3}{4}$, so $\cot^2 \theta = (\frac{3}{4})^2 = \frac{9}{16}$.
Part (ii): Same as above, $\cot^2 \theta = \frac{9}{16}$.
9/16
Q8: Verification
If $3 \cot A = 4$, check whether $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$ or not.
$\cot A = \frac{4}{3} \Rightarrow \tan A = \frac{3}{4}$.
Triangle sides: Base=4, Perp=3, Hyp=$\sqrt{3^2+4^2}=5$.
Triangle sides: Base=4, Perp=3, Hyp=$\sqrt{3^2+4^2}=5$.
LHS: $\frac{1 - (3/4)^2}{1 + (3/4)^2} = \frac{1 - 9/16}{1 + 9/16} = \frac{7/16}{25/16} = \frac{7}{25}$.
RHS: $\cos A = \frac{4}{5}, \sin A = \frac{3}{5}$.
$\cos^2 A - \sin^2 A = (\frac{4}{5})^2 - (\frac{3}{5})^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$.
$\cos^2 A - \sin^2 A = (\frac{4}{5})^2 - (\frac{3}{5})^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$.
Yes, LHS = RHS.
Q9: Compound Angles
In $\Delta ABC$, right-angled at B, if $\tan A = 1$, find the value of:
(i) $\sin A \cos C + \cos A \sin C$
(ii) $\cos A \cos C - \sin A \sin C$
(i) $\sin A \cos C + \cos A \sin C$
(ii) $\cos A \cos C - \sin A \sin C$
$\tan A = 1 = \frac{1}{1}$. So $P=1k, B=1k$. Hyp $H = \sqrt{1+1} = \sqrt{2}k$.
For A: $\sin A = \frac{1}{\sqrt{2}}, \cos A = \frac{1}{\sqrt{2}}$.
For C: Opposite is AB(1k), Adj is BC(1k).
$\sin C = \frac{1}{\sqrt{2}}, \cos C = \frac{1}{\sqrt{2}}$.
For C: Opposite is AB(1k), Adj is BC(1k).
$\sin C = \frac{1}{\sqrt{2}}, \cos C = \frac{1}{\sqrt{2}}$.
(i) $\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} + \frac{1}{2} = 1$.
(ii) $\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} - \frac{1}{2} = 0$.
(i) 1, (ii) 0
Q10: Side Relationships
In $\Delta PQR$, right-angled at Q, $PR + QR = 25$ cm and $PQ = 5$ cm. Determine the values of $\sin P, \cos P$ and $\tan P$.
Let $QR = x$. Then $PR = 25 - x$.
By Pythagoras: $PR^2 = PQ^2 + QR^2$
$(25 - x)^2 = 5^2 + x^2$
$625 + x^2 - 50x = 25 + x^2$
$600 = 50x \Rightarrow x = 12$.
$(25 - x)^2 = 5^2 + x^2$
$625 + x^2 - 50x = 25 + x^2$
$600 = 50x \Rightarrow x = 12$.
So, $QR = 12, PR = 13, PQ = 5$.
$\sin P = \frac{QR}{PR} = \frac{12}{13}$
$\cos P = \frac{PQ}{PR} = \frac{5}{13}$
$\tan P = \frac{QR}{PQ} = \frac{12}{5}$
$\cos P = \frac{PQ}{PR} = \frac{5}{13}$
$\tan P = \frac{QR}{PQ} = \frac{12}{5}$
Q11: True or False
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 12/5 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = 4/3 for some angle θ.
(i) The value of tan A is always less than 1.
(ii) sec A = 12/5 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = 4/3 for some angle θ.
(i) False. tan 60° = 1.732 > 1.
(ii) True. sec A is always ≥ 1 or ≤ -1. 12/5 = 2.4, which is valid.
(iii) False. cos A is for cosine. Cosecant is cosec A.
(iv) False. cot A is a single term representing the ratio. 'cot' alone has no meaning.
(v) False. sin θ cannot exceed 1. 4/3 = 1.33 > 1.