a right angled triangle base BC=15cm and sinB=4/5 then what is the length of hypotenuse AB
Answers
Step-by-step explanation:
=
45
(i) Calculate the measure of AB and AC.
ii) Now, if tan ∠ADC = 1; calculate the measures of CD and AD.
Also, show that: tan2B -
B
1cos2B=–1.
SOLUTION
Given
sin B =
45
i.e.
perpendicular
base
AC
AB
perpendicularbase=ACAB=45
Therefore if length of perpendicular = 4x, length of hypotenuse = 5x
Since
BC2 + AC2 = AB2 ...[ Using Pythagoras Theorem]
(5x)2 – (4x)2 = BC
BC2 = 9x2
∴ BC = 3x
Now
BC = 15
3x = 15
x = 5
(i) AC = 4x
= 4 x 5
= 20 cm
And
AB = 5x
= 5 x 5
= 25 cm
(ii) Given
tan ∠ADC =
11
i.e.
perpendicular
base
AC
CD
perpendicularbase=ACCD=11
Therefore if length of perpendicular = x, length of hypotenuse = x
Since
AC2 + CD2 = AD2 ...[Using Pythagoras Theorem]
(x)2 – (x)2 = BC
AD2 = 2x2
∴ AD = 2x
Now
AC = 20
x = 20
So
AD = 2x
= 2 x 20
= 20cm2cm
And
CD = 20 cm
Now
tan B =
AC
BC
ACBC=2015=43
cos B =
BC
AB
BCAB=1525=35
So
tan2 B –
1cos2B
=
(43)2– 1(35)2
=
169– 259
=
–99
= –
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