Q1. In \(\triangle ABC\), right-angled at B, AB = 24 cm, BC = 7 cm. Determine
:
(i) sin A, cos A
(ii) sin C, cos C
Solution:
AB = 24, BC = 7
By Pythagoras theorem
Q2. In Fig. 8.13, find tan P – cot R.
Solution:
tan P-cot R
PQ= 12,
PR = 13
QR can be found by using Pythagoras theorem
Q3. If \(\sin A = \frac{3}{4}\), calculate cos A and tan A.
Solution:
$$\begin{aligned}\sin A&=\dfrac{3}{4}\\ \sin A&=\dfrac{P}{H}\\ \dfrac{P}{H}&=\dfrac{3}{4}\\ P&=3k\quad\scriptsize\text{(k is any positive number)}\\ H&=4k\\ B^{2}&=H^{2}-P^{2}\\ B^{2}&=\left( 4k\right) ^{2}-\left( 3k\right) ^{2}\\ B&=\sqrt{16k^{2}-9k^{2}}\\ &=\sqrt{7k^{2}}\\ &=k\sqrt{7}\\\\ \cos A&=\dfrac{B}{H}\\ &=\dfrac{k\sqrt{7}}{4k}\\ &=\dfrac{\sqrt{7}}{4}\\\\ \tan A&=\dfrac{\sin A}{\cos A}\\ &=\dfrac{\frac{3}{4}}{\sqrt{7}/4}\\ &=\dfrac{3}{\sqrt{7}}\end{aligned}$$Q4. Given 15 cot A = 8, find sin A and sec A.
Solution:
$$\begin{aligned}15\times \cot A&=8\\ \cot A&=\dfrac{8}{15}\\ &=\dfrac{B}{P}\\ \dfrac{B}{P}&=\dfrac{8}{15}\\ B&=8k\quad\scriptsize\text{(k is a positive integer)}\\ P&=15k\\ H^{2}&=B^{2}+P^{2}\\ &=\left( 8k\right) ^{2}+\left( 5k\right) ^{2}\\ &=64k^{2}+225k^{2}\\ &=289k^{2}\\ H&=\sqrt{289k^{2}}\\ &=17k\\\\ \sin A&=\dfrac{P}{H}\\ &=\dfrac{15k}{17k}\\ &=\dfrac{15}{17}\\\\ \sec A&=\dfrac{H}{B}\\ &=\dfrac{17k}{8k}\\ &=\dfrac{17}{8}\end{aligned}$$Q5. Given \(\sec \theta = \frac{13}{12}\) , calculate all other trigonometric ratios.
Solution:
$$\begin{aligned}\sec \theta &=\dfrac{13}{12}\\ \sec \theta &=\dfrac{H}{B}\\ \therefore \dfrac{H}{B}&=\dfrac{13}{12}\\ H&=13k\quad\scriptsize\text{(k is a positive number)}\\ B&=12k\end{aligned}$$By pythagoras theorem
$$\begin{aligned}H^{2}&=P^{2}+B^{2}\\ P^{2}&=H^{2}-B^{2}\\ &=13^{2}-12^{2}\\ &=169-144\\ P&=\sqrt{25}\\ &=5\end{aligned}$$Trigonometrical Ratios
$$\begin{aligned}\sin \theta &=\dfrac{P}{H}\\ &=\dfrac{5}{13}\\\\ \cos \theta &=\dfrac{B}{H}\\ &=\dfrac{12}{13}\\\\ \tan \theta &=\dfrac{P}{B}\\ &=\dfrac{5}{12}\\\\ \cot \theta &=\dfrac{B}{P}\\ &=\dfrac{12}{5}\\\\ \text{cosec } \theta &=\dfrac{H}{P}\\ &=\dfrac{13}{5}\end{aligned}$$
Q6. If \(angle A\) and \(\angle B\) are acute angles such that cos A = cos B, then show that \(\angle A = \angle B\).
Solution:
$$\begin{aligned}\cos A&=\cos B\\ \cos A&=\dfrac{AC}{AB}\\ \cos B&=\dfrac{BC}{AB}\\ \Rightarrow \dfrac{AC}{AB}&=\dfrac{BC}{AB}\\ AC&=BC\\ \angle B&=\angle A\end{aligned}$$ Angles opposite to equal side of triangleQ7. If \(\cot \theta = \frac{7}{8}\), evaluate :
(i) \(\dfrac{ (1 + \sin \theta)(1 - \sin \theta) }{(1 + \cos \theta)(1 - \cos \theta)}\)
(ii) \(\cot^2 \theta\)
Solution:
$$\begin{aligned}\cot \theta &=\dfrac{7}{8}\\ &=\dfrac{B}{P}\\ \dfrac{B}{P}&=\dfrac{7}{8}\\ B&=7k\quad\scriptsize\text{(k is positive number)}\\ P&=8k\end{aligned}$$By Pythagoras theorem
$$\begin{aligned}H^{2}&=B^{2}+P^{2}\\ H^{2}&=\left( 7k\right) ^{2}+\left( 8k\right) ^{2}\\ &=49k^{2}+64k^{2}\\ &=113k^{2}\\ H&=k\sqrt{113}\end{aligned}$$ $$\begin{aligned}\sin \theta &=\dfrac{P}{H}\\ &=\dfrac{8k}{k\sqrt{113}}\\ &=\dfrac{8}{\sqrt{113}}\\\\ \cos \theta &=\dfrac{B}{H}\\ &=\dfrac{7k}{k\sqrt{113}}\end{aligned}$$(i)
Substituting values
$$\scriptsize\begin{aligned}&\dfrac{\left( 1+\sin \theta \right) \times \left( 1-\sin \theta \right) }{\left( 1+\cos \theta \right) \times \left( 1-\cos \theta \right) }\\ &=\dfrac{\left( 1+\dfrac{8}{\sqrt{113}}\right) \times \left( 1-\dfrac{8}{\sqrt{113}}\right) }{\left( 1+\dfrac{7}{\sqrt{113}}\right) \times \left( 1-\dfrac{7}{\sqrt{113}}\right) }\\ &=\dfrac{1-\left[ \dfrac{8}{\sqrt{113}}\right] ^{2}}{1-\left[ \dfrac{7}{\sqrt{113}}\right] ^{2}}\\ &=\dfrac{113-64}{113-49}\\ &=\dfrac{49}{64}\end{aligned}$$(ii)
$$\begin{aligned}\cot \theta &=\dfrac{7}{8}\\ \cot ^{2}\theta &=\left( \dfrac{7}{8}\right) ^{2}\\ \cot ^{2}\theta &=\dfrac{49}{64}\end{aligned}$$Q8. If 3 cot A = 4, check whether \(\dfrac{1-\tan^2 A}{1+ \tan^2 A}=\cos^2 A-\sin^2 A\) or not.
Solution:
3 cot A = 4
$$\begin{aligned}3\times \cot A&=4\\ &=\dfrac{4}{3}\\ &=\dfrac{B}{P}\\ \dfrac{B}{P}&=\dfrac{4}{3}\\ B&=4k\quad\scriptsize\text{(k is positive number)}\\ P&=3k\end{aligned}$$Using pythagoras theorem we can find H (Hypotenuse of triangle)
$$\begin{aligned}H^{2}&=B^{2}+P^{2}\\ &=\left( 4k\right) ^{2}+\left( 3k\right) ^{2}\\ &=16k^{2}+9k^{2}\\ &=25k^{2}\\ H&=\sqrt{25k^{2}}\\ H&=5k\end{aligned}$$ $$\begin{aligned}\cos A&=\dfrac{B}{H}\\ &=\dfrac{4k}{5k}\\ &=\dfrac{4}{5}\\\\ \cos ^{2}A&=\left( \dfrac{4}{5}\right) ^{2}\\ &=\dfrac{16}{25}\\\\ \sin A&=\dfrac{P}{H}\\ &=\dfrac{3k}{5k}\\ &=\dfrac{3}{5}\\\\ \sin ^{2}A&=\left( \dfrac{3}{5}\right) ^{2}\\ &=\dfrac{9}{25}\end{aligned}$$ $$\begin{aligned}&\cos ^{2}A-\sin ^{2}A\\ &\dfrac{16}{25}-\dfrac{9}{25}\\ &\dfrac{16-9}{25}\\ &=\dfrac{7}{25}\end{aligned}$$ $$\begin{aligned}\cot A&=\dfrac{4}{3}\\ &\dfrac{1}{\cot A}\\ &=\dfrac{3}{4}\\\\ \tan ^{2}A&=\left( \dfrac{3}{4}\right) ^{2}\\ &=\dfrac{9}{16}\end{aligned}$$ $$\begin{aligned}&\dfrac{1-\tan ^{2}A}{1+\tan ^{2}A}\\ &\dfrac{1-\dfrac{9}{16}}{1+\dfrac{9}{16}}\\ &\dfrac{\dfrac{16-9}{16}}{\dfrac{16+9}{16}}\\ &\dfrac{7}{25}\\ \therefore \dfrac{1-\tan ^{2}A}{1+\tan ^{2}A}&=\cos ^{2}A-\sin ^{2}A=\dfrac{7}{25}\end{aligned}$$Q9. In triangle ABC, right-angled at B, if tan A = \(\dfrac{1}{\sqrt3}\) find the value
of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
$$\begin{aligned}\tan A&=\dfrac{1}{\sqrt{3}}\\ \tan A&=\dfrac{P}{B}\\ \dfrac{P}{B}&=\dfrac{1}{\sqrt{3}}\\ P&=1k\\ B&=\sqrt{3}k\end{aligned}$$Using pythagoras theorem we can find Hypotenuse
$$\begin{aligned}H^{2}&=P^{2}+B^{2}\\ &=k^{2}+\left( \sqrt{3}k\right) ^{2}\\ &=k^{2}+3k^{2}\\ &=4k^{2}\\ H&=\sqrt{4k^{2}}\\ H&=2k\end{aligned}$$ $$\begin{aligned}\sin A&=\dfrac{P}{H}\\ &=\dfrac{k}{2k}\\ &=\dfrac{1}{2}\\\\ \sin C&=\dfrac{P}{H}\\ &=\dfrac{\sqrt{3}k}{2k}\\ &=\dfrac{\sqrt{3}}{2}\\\\ \cos A&=\dfrac{B}{H}\\ \cos A=\dfrac{\sqrt{3}k}{2k}\\ &\\=\dfrac{\sqrt{3}}{2}\\ \cos C&=\dfrac{B}{H}\\ &=\dfrac{k}{2k}\\ &=\dfrac{1}{2}\end{aligned}$$(i)
Sin A Cos C+ cos A Sin C
substituting values in equation
(ii)
$$\begin{aligned}&\cos A\times \cos C-\sin A\times \sin C\\ &\dfrac{\sqrt{3}}{2}\times \dfrac{1}{2}-\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}\\ &=\dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4}\\ &=0\end{aligned}$$
Q10. In \(\triangle PQR\), right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
PR + QR = 25
PQ = 5
Let QR = x
PR = 25-x
using Pythagoras theorem
Calculating Ratios sin P, cos P, tan P
$$\begin{aligned}\sin P&=\dfrac{QR}{PR}\\ &=\dfrac{12}{13}\\\\ \cos P&=\dfrac{PQ}{QR}\\ &=\dfrac{5}{13}\\\\ \tan P&=\dfrac{QR}{PQ}\\ &=\dfrac{12}{5}\end{aligned}$$Q11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) \(\sec A = \frac{12}{5}\) 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) \(\sin \theta = \frac{4}{ 3 }\) for some angle \(\theta\).
Solution:
(i) False.\(\tan A=\frac{\text{Perpendicular}}{\text{Base}}\) can be less than, equal to, or greater than 1 depending on the triangle; for example, when the perpendicular is greater than the base, \(\tan A > 1\)
(ii) True.
\(\sec A=\frac{\text{Hypotenuse}}{\text{Adjacent side}}\) ; since the hypotenuse is the largest side, this ratio can be \(\frac{12}{5}\gt 1\) for a suitable right triangle (e.g. sides proportional to 5, 12, 13).
(iii) False.
“cos A” is the standard abbreviation for cosine of angle A, while cosecant is written as “cosec A” or “csc A”, so cos A does not stand for cosecant.
(iv) False.
“cot A” means the cotangent of angle A, a single trigonometric ratio, not a product of some quantity “cot” and the number A.
(v) False.
\(\sin \theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\),and in a right triangle the hypotenuse is the longest side, so this ratio is always less than or equal to 1; it can never be \(\frac{4}{3}\gt 1\)