EXERCISE
1. In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine : (i) sin A, cos A (ii) sin C, cos C $AC^2=AB^2+BC^2$ $AC^2=24^2+7^2$ AC= 25 cm sinA= $7/25$, cos A = $24/25$ sin C = $24/25$ , cos C = $7/25$2. In Fig., find tan P – cot R. ∆PQR is a right angled triangle at Q. Hence $PR^2=PQ^2+QR^2$ $13^2=12^2+QR^2$ $QR=√(13^2-12^2)$ $QR=√(169-144)$ $QR=√25$ $QR=5$ tan P = ${QR}/{PQ}$ = $5/12$ cot R = ${QR}/{PQ}$ = $5/12$ Therefore tan P - cot R = $5/12$ - $5/12$ = 03. If sin A =$3/4$ ,calculate cos A and tan A. Given sin A = $3/4$ , $.^{.}.$ cos A = $√(1-sin^2A)$ = $√(1-(3/4)^2)$ = $√(1-(9/16))$ = $√((16-9)/16)$ = $√7/4$ tan A = ${sinA}/{cosA}$ = ${3/4}/{√7/4}$ = > ${3/√7}$4. Given 15 cot A = 8, find sin A and sec A. Given cot A = $8/15$ i.e $B/P$ Therefore H = $√(B^2+P^2)$ = $√(8^2+15^2)$ = $√(64+225)$ = $√289$ = $17$ sin A = $P/H$ = $15/17$ and sec A = $H/B$ =$17/8$5. Given sec θ = $13/12$, calculate all other trigonometric ratios. Given sec θ = $13/12$ i.e $H/B$ therefore $H=13$ , $B=12$ and P = $√(13^2-12^2)$ => $√25$ => 5 Trigonometric ratios are : sin A =$P/H$ = $5/13$ cos A = $B/H$ = $12/13$ tan A = $P/B$ = $5/12$ cosec A = $H/P$ = $13/5$ sec A = $H/B$ = $13/12$ cot A = $B/P$ = $12/5$6. If ∠ A and ∠ B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B. Let ABC be a triangle , right angled at C. Given cos A = cos B. ie. cos A = ${AC}/{AB}$ and cos B = ${BC}/{AB}$ therefore AC = BC Therefore ABC is an isoceles triangle with sides AB=AC , therefore ∠ A = ∠ B .7. If cot θ = $7/8$, evaluate : (i) ${(1 + sin θ) (1 − sin θ)}/{(1 + cos θ)(1 − cos θ)}$ , (ii)cot2 θ cot θ = $7/8$ = $B/P$ i.e B = 7 , P = 8 , so, H = $√(7^2+8^2)$ => $√(49+64)$ => $√(113)$ sin θ = $P/H$ = $8/√113$ cos θ = $B/H$ = $7/√113$ (i) ${(1 + sin θ) (1 − sin θ)}/{(1 + cos θ)(1 − cos θ)}$ = ${(1 + 8/√113) (1 − 8/√113)}/{(1 + 7/√113)(1 − 7/√113)}$ = ${(1 − 64/113)}/{(1 − 49/113)}$ = ${(49/113)}/{(64/113)}$ = ${49}/{64}$ (ii) $cot^2θ$ = $(7/8)^2$ = $49/64$8. If 3 cot A = 4, check whether ${1 − tan^2 A}/{1 + tan^2A}$ = $cos^2A – sin^2A$ or not. given cot A = $4/3$ = $B/P$ Therefore H = $√(3^2+4^2)$ = $√25$ = $5$ L.H.S = ${1 − tan^2 A}/{1 + tan^2A}$ $tan A$ = $1/{cot A}$ = $3/4$ = ${1 − (3/4)^2}/{1 + (3/4)^2}$ = ${1 − (9/16)}/{1 + (9/16)}$ = ${(16 − 9)/16}/{(16 + 9)/16}$ = $7/25$ R.H.S = $cos^2A – sin^2A$ sin A = $P/H$ = $3/5$ cos A = $B/H$ = $4/5$ $.^{.}.$ $cos^2A – sin^2A$ = $(4/5)^2-(3/5)^2$ = $16/25 - 9/25$ =$7/25$ Since L.H.S is equal to R.H.S , therefore ${1 − tan^2 A}/{1 + tan^2A}$ is equal to $cos^2A – sin^2A$9. In triangle ABC, right-angled at B, if tan A = $1/√3$; find the value of: (i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C Triangle ABC is right-angled at B and tan A = $1/√3$ => ${BC}/{AB}$ Therefore $AC$ = $√{AB^2+BC^2}$ = $√{3+1}$ = $√4$ = 2 Therefore sin A =${BC}/{AC}$ =$1/2$ cos A = ${AB}/{AC}$ = $√3/2$ sin C = ${AB}/{AC}$ = $√3/2$ cos C = ${BC}/{AC}$ = $1/2$ (i) sin A cos C + cos A sin C = $(1/2)(1/2) + (√3/2)(√3/2)$ = $1/4 + 3/4$ = $4/4$ =1 (ii) cos A cos C – sin A sin C = $(√3/2)(1/2) - (1/2)(√3/2)$ = $√3/4$ - $√3/4$ =010. In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P. Given, ∆ PQR is a right-angled at Q , also given PR + QR = 25 cm and PQ = 5 cm , therefore $PR^2=PQ^2+QR^2$ $PR^2 - QR^2=5^2$ $(PR-QR)(PR+QR)=25$ $(PR-QR)(25)=25$ $(PR-QR)=1$ therefore $2PR = 25 +1$ => $PR=26/2$ = $13$ and QR = $PR-1$ = $13-1$ = $12$ Therefore , sin P =${QR}/{PR}$ = $12/13$ cos P = ${PQ}/{PR}$ = $5/13$ tan P = ${QR}/{PQ}$ = $12/5$11. 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) False, because , tan A = {opposite side of angle A}$/${Adjacent side of angle A} , if opposite side > adjacent side , then the value of tan A is greater than 1. (ii) true, (iii) false (iv) false (v) false