Question

In triangle ABC, AB=5cm, BC=12cm,<B=90°
a)find SinA, CosA, tanA
b) Prove that Sin^2A+Cos^2A=,1​

Answer 1

Given :

• AB = 5 cm
• BC = 12 cm
• angle(B) = 90°

To find :

a) sinA , cosA , tanA

b) Prove , sin²A + cos²A = 1

Image :

$\setlength{\unitlength}{1cm}\begin{picture}{\thicklines}\qbezier(0,0)(0,0)(6.5,0)\qbezier(0,0)(0,0)(0,2.5)\qbezier(0,2.5)(0,2.5)(6.5,0)\put(-0.3,2.7){\bf A}\put(-0.4,-0.4){\bf B}\put(6.7,-0.4){\bf C}\put(0,0.12){\framebox}\put(0.3,0.2){\bf 90}\put(2.5,-0.5){\bf 12cm}\put(-1,1.3){\bf 5cm}\end{picture}$

{ App user refer to attachment for same }

Solution :

First of all we need to find hypoteneus ( AC)

Using Pythagoras theorem,

AC² = AB² + BC²

AC² = (12)² + (5)²

AC² = 144 + 25

AC² = 169

AC = √169

AC = ± 13

{ as side can't be negative }

AC = 13 cm

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Part a)

sinA = Perpendicular ÷ Hypoteneus

sinA = BC ÷ AC

sinA = 12/13

cosA = Base ÷ hypoteneus

cosA = AB ÷ AC

cosA = 5/13

tanA = Perpendicular ÷ base

tanA = BC ÷ AC

tanA = 12/5

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Part b)

To prove sin²A + cos²A = 1

LHS ,

$\sf \implies \: { \bigg \{\dfrac{12}{13} \bigg \} }^{2} + \: { \bigg \{ \dfrac{5}{12} \bigg \} }^{2} \\ \\ \sf \implies \: \dfrac{144}{169} + \dfrac{25}{169} \\ \\ \sf \implies \: \dfrac{144 + 25}{169} \\ \\ \sf \implies \: \dfrac{169}{169} \\ \\ \sf \implies \: 1$

RHS = 1

This gives, RHS = LHS ,

HENCE PROVED

Answer 2

Step-by-step explanation:

right angled ∆ ABC,

BY PYTHAGORAS THEORM,

AC² = AB²+ BC²

AC² = 5²+12²

AC²= 25+144

AC²= 169

AC=√169

AC= 13

(a). sin A = P/H

BC/ AC

12/ 13

cosA. B/H

AB/AC

5/ 13

tanA. P/B

BC/AB

12/5

(b). sin²A + cos²A = 1

LHS= ( 12/13)² + (5/13)²

= 144/169+ 25/169

= 169/169

= 1

And RHS = 1

SO LHS= RHS

HENCE PROVED