Question

In the given figure, triangle ABC is right-angled at B. D is the foot of the perpendicular from B to AC. Given that BC = 3 cm and AB = 4 cm.

find :

1. tan ∠DBC
2. sin ∠DBA

Answer 1

Since the triangle ABC is a right-angled triangle, so using the Pythagorean Theorem,

AC² = AB² + BC²

AC² = 4² + 3²

AC² = 16 + 9

AC² = 25

AC = `sqrt25`

AC = 5

In ΔBCD

Cd² + BD² = BC²

y² + x² = (3)²

y² + x²  = 9        (1)

In ΔABD

AD² + BD² = 16

(5 - y)² + x² = 16         ...(2)

subtracting (2) from (1) we get

(5 - y)² - y² = 7

25 + y² - 10y - y² = 7

18 =10y

1.8 = y

CD = 1.8

AD = 5 - 1.8

CD = 1.8, AD = 3.2, BD = 2.4

Now,

(1.8)² + x² = 9

x² = 9 - 3.24

x² = 5.76

x² = 2.4

BD = 2.4

1. tan ∠DBC = `(CD)/(BD) = 1.8/2.4 = 3/4`
2. sin ∠DBA = `(AD)/(AB) = 3.2/4 = 4/5`