Question

in triangle abc right angled at b if tan a is equal to 4/3 then the value of cos c​

Answer 1

Answer:

In △ABC,

∠B=90

o

, tanA=

3

1

=

AB

BC

Let BC =1x,AB=

3

x

AC

2

=AB

2

+BC

2

AC

2

=(

3

x)

2

+(x)

2

=4x

2

AC=2x

(i) sinAcosC+cosAsinC=

2

1

×

2

1

+

2

3

×

2

3

=

4

1

+

4

3

=1

(ii) cosAcosC−sinAsinC=

2

3

×

2

1

−

2

1

×

2

3

=

4

3

−

4

3

=0

Answer 2

Given:

in triangle abc right angled at b,$\[\tan a = \frac{4}{3}\]$  then the value of cos c​

Solution:

Know that,

$\[\tan \theta = \frac{{{\rm{Perpendicular}}}}{{{\rm{Base}}}}\]$ and ,

$\[{\mathop{\rm Cos}\nolimits} \theta = \frac{{{\rm{Base}}}}{{{\rm{Hypotenuse}}}}\]$

Evaluate hypotenuse,

$\[\tan a = \frac{{{\rm{Perpendicular}}}}{{{\rm{Base}}}} = \frac{4}{3}\]$

$\[\begin{array}{l}\therefore H = \sqrt {{4^2} + {3^2}} \\ \Rightarrow H = \sqrt {16 + 9} \\ \Rightarrow H = \sqrt {25} \\ \Rightarrow H = 5\end{array}\]$

Evaluate $\[\cos c\]$,

Refer the figure given below,

Notice that, now BC is consider as base since angle is $c$,

$\[\begin{array}{l}\cos c = \frac{{BC}}{{AC}}\\ \Rightarrow \cos c = \frac{4}{5}\end{array}\]$

Hence, value of $\[\cos c = \frac{4}{5}\]$.