In triangle $A B C$ , $A D$ is perpendicular to $B C$ . $sin B = 0.8$ , $B D = 9$ cm and $tan C = 1$ . Find the length of $A B$

Open in App

Solution

In $△ A B C$ , $A D ⊥ B C$ , $sin B = 0.8$ , $tan C = 1$ , $B D = 9$
In $△ A B D$ ,
$sin B = 0.8$
${cos}^{2} B = 1 - {sin}^{2} B = 1 - (0.8)^{2}$
$cos B = (0.6)^{2}$
$cos B = 0.6$
$cos B = \frac{B D}{A B}$
$A B = \frac{B D}{0.6}$
$A B = \frac{9}{0.6} = 15$
Also, using Pythagoras theorem,
$A B^{2} = A D^{2} + B D^{2}$
$15^{2} = 9^{2} + A D^{2}$
$A D = 144$
$A D = 12$ cm
Now, In $△ A C D$
$tan C = 1 = \frac{P}{B} = \frac{A D}{C D}$
$A D = C D = 12$ cm
Using Pythagoras theorem,
$A D^{2} + C D^{2} = A C^{2}$
$12^{2} + 12^{2} = A C^{2}$
$A C = 12 \sqrt{2}$ cm

Suggest Corrections

Similar questions

Q. In triangle

A B C

A D

is perpendicular to

B C

sin B = 0.8, B D = 9

cm and

tan C = 1

. Find the length of

A B, A D, A C

and

D C

Q. AD is the perpendicular to BC in a triangle ABC. BC = 10 cm, Find BD.

$△ A B C$ is right-angled at $A . A D$ is perpendicular to $B C, A B = 5 c m, B C = 13 c m$ and $A C = 12 c m$ . Find the area of $△ A B C$ . Also, find the length of $A D$ .

$△ A B C$ is right angled at $A$ . $A D$ is perpendicular to $B C$ . If $A B = 5 c m, B C = 13 c m$ and $A C = 12 c m$ . Find the area of $△ A B C$ . Also find the length of $A D$ .