Question

$\sf\mapsto\: Please \: help!!!$
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Answer 1

Given : A triangle DEF where right angle at E, length of ED and FD are √3k and 2k respectively. Further given that cos D = √3/2

To find : -

[1] The length of FE

[2] The value of tan F

[3] The value of sin D

Solution :-

[1] To find the length of side FE, we will use pythagoras theorem according to which,

• Hypo.² = Base² + Perpendicular²

Therefore, by applying pythagoras theorem in ∆DEF , we get:

⇒ EF² + ED² = FD²

⇒ EF² = FD² - ED²

⇒ EF = √[(2k)² - (√3k)²]

⇒ EF = √[4k² - 3k²]

⇒ EF = √k²

⇒ EF = k

Therefore the length of FE = k

[2] Now, to find the value of tan F, we must know the formula of tan θ.

• tan θ = Perpendicular/base

So,

⇒ tan F = DE/EF

⇒ tan F = √3k/k

⇒ tan F = √3

Therefore the value of tan F = √3

[3] Now, to find the value of sin D, we must know the formula of sin θ.

• sin θ = Perpendicular/hypotenuse

So,

⇒ sin D = EF/DF

⇒ sin D = k/2k

⇒ sin D = 1/2

Therefore the value of sin D = 1/2

Additional Information :-

• tan ϕ = sin ϕ/cos ϕ
• cosec ϕ = 1/sin ϕ
• sec ϕ = 1/cos ϕ
• cot ϕ = 1/tan ϕ
• cot ϕ = cos ϕ / sin ϕ
• sin² ϕ + cos² ϕ = 1
• cosec² ϕ - cot² ϕ = 1
• sec² ϕ - tan² ϕ = 1