Question

in triangle ABC AD PERPENDICULAR TO CB AC=29 UNITS DC=21 UNITS AND AB=40 UNITS FIND THE VALUE OF cos^B-sin^2B

Answer 1

Answer:

In triangle ADC ,

ADC=90

Ac2=dc2+ad2

Ad2=(29)2 -(21)2=841-441=400

Ad=20

In triangle adb

Ab2=ad2+bd2

Bd2= ab2-ad2=(40)2-(20)2=1600-400=1200

Bd=√1200

Cosb=bd/ab=√1200/40=20√3/40=√3/2

Sinb=ad/ab=20/40=1/2

Lhs=(cosb)2-(sinb)2

= (√3/2)2-(1/2)2=3/4-1/4=2/4=1/2

Value is 1/2.

Answer 2

Answer:

Step-by-step explanation:

1/2 is the answer. Since

CosB=20√3/40

CosB=√3/2

Thus, CosB=Cos30°

Cos^2B-Sin^2B

(√3/2)^2-(1/2)^2

1/2