In the adjoining figure, ABC is a triangle in which ∠B = 45° and ∠C = 60°. If AD ⟂ BC and BC = 8 m, find the length of the altitude AD.
Answer :
ABC
∠B = 45° and ∠C = 60°
AD ⟂ BC and BC = 8 m
In right △ ABD
tan 45° = AD/BD
1 = AD/BD
AD = BD
In right △ ACD
tan 60° = AD/DC
√3 = AD/DC
⇒ DC = AD/√3
BD + DC = AD + AD/√3
BC = (√3AD + AD)/ √3
⇒ 8 = [AD (√3 + 1)]/ √3
AD = 8√3/(√3 + 1)
= 4 (3 – √3) m
More Solutions:
- cos 65°/sin 25° + cos 32°/sin 5
- Express each of the following in terms.
- sin2 28° – cos2 62° = 0
- sin 63° cos 27° + cos 63° sin 27° = 1
- sec 70° sin 20° – cos 20° cosec 70° = 0
- cot 54°/tan 36° + tan 20°/cot 70° – 2 = 0
- cos 80°/sin 10° + cos 59° cosec 31° = 2.
- Without using trigonometrical tables, evaluate:
- Prove the following: cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1
- Simplify the following: It can be written as