Question

1) From the given figure, find :(i) Sin A and (ii) Cos C (iii) sin A - cos C (3M] 3 B​

Answer 1

Answer:

From right angled triangle ABC,

By Pythagoras theorem, we get

BC

2

=AC

2

+AB

2

AC

2

=BC

2

−AB

2

AC

2

=(10)

2

−(6)

2

AC

2

=100−36

AC

2

=64

AC

2

=8

2

AC= 8

(i) sin B = perpendicular/ hypotenuse

= AC/BC

= 8/10

= 4/5

(ii) cos C = Base/hypotenuse

= AC/BC

= 8/10

= 4/5

(iii) sin B = Perpendicular/hypotenuse

= AC/BC

= 8/10

= 4/5

Sin C = perpendicular/hypotenuse

= AB/BC

= 6/10

= 3/5

Now,

Sin B + sin C = (4/5) + (3/5)

= (4 + 3)/5

= 7/5

(iv) sin B = 4/5

Cos C = 4/5

Sin C = perpendicular/ hypotenuse

= AB/BC

= 6/10

= 3/5

Cos B = Base/Hypotenuse

= AB/BC

= 6/10

= 3/5

sin B cos C + sin C cos B

=(4/5)×(4/5)×(3/5)×(3/5)

= (26/25)×(9/25)

= (16+9)/25

= 25/25

= 1

From Figure

AC = 13, CD = 5, BC =21,

BD = BC – CD

= 21 – 5

= 16

Step-by-step explanation:

Answer 2

From right angled triangle ABC,
By Pythagoras theorem, we get
BC
2
=AC
2
+AB
2

AC
2
=BC
2
−AB
2

AC
2
=(10)
2
−(6)
2

AC
2
=100−36
AC
2
=64
AC
2
=8
2

AC= 8
(i) sin B = perpendicular/ hypotenuse
= AC/BC
= 8/10
= 4/5
(ii) cos C = Base/hypotenuse
= AC/BC
= 8/10
= 4/5
(iii) sin B = Perpendicular/hypotenuse
= AC/BC
= 8/10
= 4/5
Sin C = perpendicular/hypotenuse
= AB/BC
= 6/10
= 3/5
Now,
Sin B + sin C = (4/5) + (3/5)
= (4 + 3)/5
= 7/5
(iv) sin B = 4/5
Cos C = 4/5
Sin C = perpendicular/ hypotenuse
= AB/BC
= 6/10
= 3/5
Cos B = Base/Hypotenuse
= AB/BC
= 6/10
= 3/5
sin B cos C + sin C cos B
=(4/5)×(4/5)×(3/5)×(3/5)
= (26/25)×(9/25)
= (16+9)/25
= 25/25
= 1
From Figure
AC = 13, CD = 5, BC =21,
BD = BC – CD
= 21 – 5
= 16