3
if sin A = $\dfrac{3}{4}$, calculate cos A and tan A.
Solution
sin A = $\dfrac{3}{4}$
As sin A = $\dfrac{\text{opposite side}}{\text{hypotenuse}}$ ⇒ opposite side = 3 and hypotenuse = 4
Using Pythagoras theorem, adjacent side = $\sqrt{4^2 - 3^2}$ = $\sqrt{7}$
cos A = $\dfrac{\text{adjacent side}}{\text{hypotenuse}}$ = $\dfrac{\sqrt{7}}{4}$
tan A = $\dfrac{\text{oppsite side}}{\text{adjacent side}}$ = $\dfrac{3}{\sqrt{7}}$
4
Given 15 cot A = 8, find sin A and sec A.
Solution
15 cot A = 8 ⇒ cot A = $\dfrac{8}{15}$
cot A = $\dfrac{\text{adjacent side}}{\text{opposite side}}$
⇒ adjacent side = 8 and opposite side = 15
Using Pythagoras theorem, hypotenuse = $\sqrt{8^2 + 15^2}$ = $17$
sin A = $\dfrac{\text{oppsite side}}{\text{hypotenuse}}$ = $\dfrac{15}{17}$
sec A = $\dfrac{\text{hypotenuse}}{\text{adjacent side}}$ = $\dfrac{17}{8}$
5
Given sec θ = $\dfrac{13}{12}$, calculate all other trigonometric ratios.
Solution
sec A = $\dfrac{\text{hypotenuse}}{\text{adjacent side}}$ ⇒ hypotenuse = 13 and adjacent side = 12
Using Pythagoras theorem, opposite side = $\sqrt{13^2 - 12^2}$ = $5$
sin A = $\dfrac{\text{oppsite side}}{\text{hypotenuse}}$ = $\dfrac{5}{13}$
cos A = $\dfrac{\text{adjacent side}}{\text{hypotenuse}}$ = $\dfrac{12}{13}$
tan A = $\dfrac{\text{opposite side}}{\text{adjacent side}}$ = $\dfrac{5}{12}$
cosec A = $\dfrac{\text{hypotenuse}}{\text{oppsite side}}$ = $\dfrac{13}{5}$
sec A = $\dfrac{\text{hypotenuse}}{\text{adjacent side}}$ = $\dfrac{13}{12}$
cot A = $\dfrac{\text{adjacent side}}{\text{opposite side}}$ = $\dfrac{12}{5}$
6
If ∠ A and ∠ B are acute angles such that cos A = cos B, then show that ∠ A and ∠ B.
Solution
cos A = $\dfrac{\text{adjacent side}}{\text{hypotenuse}}$ = $\dfrac{AC}{AB}$
cos B = $\dfrac{\text{adjacent side}}{\text{hypotenuse}}$ = $\dfrac{BC}{AB}$
cos A = cos B ⇒ $\dfrac{AC}{AB}$ = $\dfrac{BC}{AB}$ ⇒ AC = BC
In a triangle, angles opposite to equal sides are equal ⇒ ∠ A and ∠ B
7
If cot θ = $\dfrac{7}{8}$, evaluate:
(i) $\dfrac{(1 + sin θ)(1 - sin θ)}{(1 + cos θ)(1 - cos θ)}$
(ii) $cot^2θ$
Solution
cot θ = $\dfrac{\text{adjacent side}}{\text{opposite side}}$ = $\dfrac{7}{8}$
⇒ adjacent side = 7 and opposite side = 8
Hypotenuse = $\sqrt{7^2 + 8^2}$ = $\sqrt{113}$
(i)
sin θ = $\dfrac{\text{oppsite side}}{\text{hypotenuse}}$ = $\dfrac{8}{\sqrt{113}}$
sin θ = $\dfrac{\text{adjacent side}}{\text{hypotenuse}}$ = $\dfrac{7}{\sqrt{113}}$
$\dfrac{(1 + sin θ)(1 - sin θ)}{(1 + cos θ)(1 - cos θ)}$
= $\dfrac{\left(1 + \dfrac{8}{\sqrt{113}}\right)\left(1 - \dfrac{8}{\sqrt{113}}\right)}{\left(1 + \dfrac{7}{\sqrt{113}}\right)\left(1 - \dfrac{7}{\sqrt{113}}\right)}$
= $\dfrac{\left(1 - \dfrac{64}{113}\right)}{\left(1 - \dfrac{49}{113}\right)}$ = $\dfrac{49}{64}$
(ii) $cot^2θ$ = $\dfrac{7}{8}$ × $\dfrac{7}{8}$ = $\dfrac{49}{64}$
8
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 ⇒ cot A = $\dfrac{4}{3}$ ⇒ tan A = $\dfrac{3}{4}$
Opposite side = 3 and Adjacent side = 4 ⇒ Hypotenuse = $\sqrt{3^2 + 4^2}$ = 5
∴ sin A = $\dfrac{3}{5}$ and cos A = $\dfrac{4}{5}$
$\dfrac{1- tan^2 A}{1 + tan^2 A}$ = $\dfrac{1 - \left(\dfrac{3}{4}\right)^2}{1 + \left(\dfrac{3}{4}\right)^2}$
= $\dfrac{\dfrac{7}{16}}{\dfrac{25}{16}}$ = $\dfrac{7}{25}$
$cos^2 A - sin^2 A$ = $\left(\dfrac{4}{5}\right)^2 - \left(\dfrac{3}{5}\right)^2$ = $\dfrac{7}{25}$
Hence $\dfrac{1- tan^2 A}{1 + tan^2 A}$ = $cos^2 A - sin^2 A$
9
In triangle ABC, right-angled at B, if tan A = $\dfrac{1}{\sqrt{3}}$, find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C - sin A sin C
Solution
tan A = $\dfrac{1}{\sqrt{3}}$ ⇒ $\dfrac{\text{side opposite to A}}{\text{side adjacent to A}}$ = $\dfrac{1}{\sqrt{3}}$
∴ $\frac{\text{BC}}{\text{AB}} = \dfrac{1}{\sqrt{3}}$ ⇒ BC × $\sqrt{3} = AB$
AC = $\sqrt{AB^2 + BC^2}$ = $\sqrt{3BC^2 + BC^2}$ = $\sqrt{4BC^2}$ = 2BC
sin A = = $\dfrac{\text{oppsite side}}{\text{hypotenuse}}$ = $\dfrac{BC}{AC}$ = $\dfrac{BC}{2BC}$ = $\dfrac{1}{2}$
cos A = = $\dfrac{\text{adjacent side}}{\text{hypotenuse}}$ = $\dfrac{AB}{AC}$ = $\dfrac{\sqrt{3}BC}{2BC}$ = $\dfrac{\sqrt{3}}{2}$
sin C = = $\dfrac{\text{oppsite side}}{\text{hypotenuse}}$ = $\dfrac{AB}{AC}$ = $\dfrac{\sqrt{3}BC}{2BC}$ = $\dfrac{\sqrt{3}}{2}$
cos C = = $\dfrac{\text{adjacent side}}{\text{hypotenuse}}$ = $\dfrac{BC}{AC}$ = $\dfrac{BC}{2BC}$ = $\dfrac{1}{2}$
(i) sin A cos C + cos A sin C
= $\dfrac{1}{2} × \dfrac{1}{2}$ + $\dfrac{\sqrt{3}}{2} × \dfrac{\sqrt{3}}{2}$
= $\dfrac{1}{4} + \dfrac{3}{4}$ = 1
(ii) cos A cos C - sin A sin C
= $\dfrac{\sqrt{3}}{2} × \dfrac{1}{2}$ - $\dfrac{{1}}{2} × \dfrac{\sqrt{3}}{2}$
= $\dfrac{\sqrt{3}}{4} × \dfrac{\sqrt{3}}{4}$ = 0
10
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