Question

INTRODUCTION TO TRIGONOMETRY
EXERCISE 8.1
1. In A ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, COS A
(ii) sin C, cos C
2. In Fig. 8.13, find tan P-cot R.
12 cnt
3
3. If sin A =
calculate cos A and tan A.
4
4. Given 15 cot A= 8, find sin A and sec A.​

Answer 1

answer is in the attachment......

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Answer 2

Step-by-step explanation:

$\boxed{1}$

In ΔABC , right angled at B , AB = 24cm , BC = 7 cm.

We know that,

$AC²=AB²+BC²$

$=24²+7²$

$=576+49$

$=625$

$AC=\sqrt{625}$

$=2cm$

(i)$sin A =\frac{opposite ∠A}{Hypotense}$

$=\frac{AB}{AC}$

$=\frac{7}{25}$

$cos A=\frac{adjacent ∠A}{Hypotenuse}$

$=\frac{AB}{AC}$

$=\frac{24}{25}$

(ii)$sin C=\frac{opposite ∠C}{Hypotenuse}$

$=\frac{AB}{AC}$

$=\frac{24}{25}$

$cos C=\frac{adjacent ∠C}{Hypotenuse}$

$=\frac{AB}{AC}$

$=\frac{7}{25}$

$\boxed{2}$

We know that,

By Pythagoras theorem,

$PR²=QR²+PQ²$

$QR²=PR²+PQ²$

$QR=\sqrt{PR²-PQ²}$

$=\sqrt{13²-12²}$

$=\sqrt{169-144}$

$=\sqrt{25}=5cm$

$tan P=\frac{opposite ∠P}{adjacent ∠P}$

$=\frac{QR}{PQ}$

$=\frac{5}{12}$

$cot R=\frac{adjacent∠R}{opposite∠R}$

$=\frac{QR}{PQ}$

$=\frac{5}{12}$

$tan P-cot R =\frac{5}{12}-\frac{5}{12}=0$

$\boxed{3}$

$sin A=\frac{3}{4}$

We know that,

$sin C=\frac{perpendicular}{Hypotenuse}$

$Hypotenuse²=Perpendicular²+Base²$

$H²=P²+B²$

$sin A=\frac{3}{4}=\frac{P}{H}$

$P=3,H=4$

$B=\sqrt{H²-P²}$

$=\sqrt{4²-3²}$

$=\sqrt{16-9}$

$Base=\sqrt{7}$

$cos A=\frac{Base}{Hypotenuse}={\frac{7}}{4}$

$=\sqrt{7}/{4}$

$tan A=\frac{Perpendicular}{Base}=\frac{3}/\sqrt{7}$

$\boxed{4}$

$15 cot A=8$

$cot A=\frac{8}{15}$

$cot A =\frac{Base}{Perpendicular}$

$Base=8,Perpendicular=15$

$Hypotenuse=\sqrt{perpendicular²+Base²}$

$=\sqrt{15²+8²}$

$=\sqrt{225+64}$

$=\sqrt{289}$

$=17$

$sin A=\frac{Perpendicular}{Hypotenuse}=\frac{15}{17}$

$sec A=\frac{Hypotenuse}{Base}=\frac{17}{8}$

Hope it helps :-)

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