**In the adjoining figure, AB = 4 m and ED = 3 m.**

**If sin α = 3/5 and cos β = 12/13, find the length of BD.**

**Answer :**

sin α = AB/AC = 3/5

AB = 3 and AC = 5

AC^{2} = AB^{2} + BC^{2}

5^{2} = 3^{2} + BC^{2}

25 = 9 + BC^{2}

⇒ BC^{2} = 25 – 9 = 16

BC^{2} = 4^{2}

⇒ BC = 4

tan α = AB/BC = 4/5

cos β = CD/CE = 12/13

CD = 12 and CE = 13

CE^{2} = CD^{2} + ED^{2}

13^{2} = 12^{2} + ED^{2}

ED^{2} = 13^{2} – 12^{2}

⇒ ED^{2} = 169 – 144 = 25

ED^{2} = (5)^{2}

⇒ ED = 5

⇒ tan β = ED/CD = 5/12

tan α = AB/BC = 4/BC

¾ = 4/BC

⇒ BC = (4×4)/3 = 16/3 m

tan β = ED/CD = 3/CD

⇒ 5/12 = 3/CD

CD = (12 × 3)/5 = 36/5 m

Here,

BD = BC + CD

= 16/3 + 36/5

Taking LCM

= (80 + 108)/15

= 188/15 m

= 12 8/15 m

**More Solutions:**

- Show that sin (A + B) ≠ sin A + sin B.
- If A = 60° and B = 30°, verify that
- Find the value of θ.
- Find the value of 2 tan2 θ + sin2 θ – 1.
- From the adjoining figure, find
- If 3θ is an acute angle
- If tan 3x = sin 45° cos 45° + sin 30
- If 4 cos2 x° – 1 = 0 and 0 ≤ x ≤ 90, find
- If sin 3x = 1 and 0° ≤ 3x ≤ 90°
- Find cos 2θ, given that θ is acute.