A geometry problem by Yahia El Haw

Geometry Level 2

From a hot air balloon, the angle between a radio antenna straight below and the base of the library downtown is 57°, as shown below. If the distance between the radio antenna and the library is 1.3 miles, how many miles high is the balloon?

1.3 sin ⁡ 57 ∘ \large\frac { 1.3 }{ \sin { { 57 }^{ \circ } } } 1.3 tan ⁡ 57 ∘ \large1.3\tan { { 57 }^{ \circ } } 1.3 sin ⁡ 57 ∘ \large1.3\sin { { 57 }^{ \circ } } 1.3 tan ⁡ 57 ∘ \large\frac { 1.3 }{ \tan { { 57 }^{ \circ } } } 1.3 cos ⁡ 57 ∘ \large\frac { 1.3 }{ \cos { { 57 }^{ \circ } } }

Yahia El Haw by Yahia El Haw , 18, Egypt

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2 solutions

Let x x be the height of the balloon, then we have

tan ⁡ 57 = opposite side adjacent side = 1.3 x \tan 57=\dfrac{\text{opposite side}}{\text{adjacent side}}=\dfrac{1.3}{x} or x = 1.3 tan ⁡ 57 x=\boxed{\dfrac{1.3}{\tan 57}} .

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S O H − C A H − T O A SOH-CAH-TOA

S O H : sin ⁡ θ = o p p o s i t e h y p o t e n u s e SOH:\sin\theta=\dfrac{opposite}{hypotenuse}

C A H : cos ⁡ θ = a d j a c e n t h y p o t e n u s e CAH:\cos\theta=\dfrac{adjacent}{hypotenuse}

T O A : = tan ⁡ θ = o p p o s i t e a d j a c e n t TOA:=\tan\theta=\dfrac{opposite}{adjacent}

tan ⁡ 57 = 1.3 h \tan57=\dfrac{1.3}{h}

h = 1.3 tan ⁡ 57 h=\dfrac{1.3}{\tan57}

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