Question

pleeeeeease answer these questions
question nos example 1,2,3and 7,8 questions
it's urgent please

Answer 1

1)
In triangle ABC, AB=8,BC=15,AC=17
Therefore AC is hypotenuse and angle B=90
SinA=opposite/hypotenuse
SinA=15/17
CosA=adjacent/hypotenuse
CosA=8/17
TanA=Opposite/adjacent
TanA=15/8


2)
In triangle PQR,angleP=90,PQ=7,QR=25
By pythagoras theorem we get,PR=24
TanQ=opposite/adjacent
TanQ=24/7
TanR=7/24
TanQ-TanR=24/7-7/24=(576-49)/168
=527/168

3)
In triangle ABC,angle B=90,a=24,b=25,
angleBÀC=theta
Therefore AC=25,BC=24
by pythagoras theorem we get,AB=7
Cos(theta)=adjacent/hypotenuse
Cos(theta)=AB/AC
Cos(theta)=7/25
Tan(theta)=opposite/adjacent
Tan(theta)=BC/AB
Tan(theta)=24/7

7)
Cot(theta)=7/8=adjacent/opposite
By Pythagoras theorem we get,
hypotenuse=√113
sin(theta)=opposite/hypotenuse
Sin(theta)=8/√113
Cos(theta)=adjacent/hypotenuse
Cos(theta)=7/√113
i)
={[1+sin(theta)][1-sin(theta)]}/{[1-cos(theta)]}{[1+cos(theta)]}
={1+8/√113}{1-8/√113}/{1-7/√113}{1+7/√113}
={1-(8/√113)^2]/{1-(7/√113)^2}
=49/64
ii)
={1+sin(theta)}/cos(theta)
={1+8/√113}/7/√113
=(√113+8)/7

8)
In triangle ABC,angle B=90,
tanA=√3/1=opposite/adjacent=BC/AB
BC=√3,AB=1
by Pythagoras theorem we get,
AC=2
SinA=opposite/hypotenuse=BC/AC=√3/2
SinC=opposite/hypotenuse=AB/AC=1/2
CosA=adjacent/hypotenuse=AB/AC=1/2
CosC=adjacent/hypotenuse=BC/AC=√3/2

i)
=sinAcosC+cosAsinC
=√3/2×√3/2+1/2×1/2
=3/4+1/4
=4/4
=1
ii)
=cosAcosC-sinAsinC
=1/2×√3/2-√3/2×1/2
=√3/4-√3/4
=0