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Find the height of an isosceles triangle with an area of 40 and a base of 5

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sin(4θ)−√32 =0, ∀ 0≤ θ<2π

Solution

θ=π12 , θ=π6 , θ=7π12 , θ=2π3 , θ=13π12 , θ=7π6 , θ=19π12 , θ=5π3

Degrees

θ=15^◦, θ=30^◦, θ=105^◦, θ=120^◦, θ=195^◦, θ=210^◦, θ=285^◦, θ=300^◦

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sin(4θ)−√32 =0, ∀ 0≤ θ<2π

Move √32 to the right side

sin(4θ)=√32

General solutions for sin(4θ)=√32

4θ=π3 +2πn, 4θ=2π3 +2πn

Solve 4θ=π3 +2πn: θ=π12 +πn2

Solve 4θ=2π3 +2πn: θ=π6 +πn2

θ=π12 +πn2 , θ=π6 +πn2

Solutions for the range 0≤ θ<2π

θ=π12 , θ=π6 , θ=7π12 , θ=2π3 , θ=13π12 , θ=7π6 , θ=19π12 , θ=5π3

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

Find the height of an isosceles triangle with an area of 40 and a base of 5

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