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Dovator [93]
3 years ago
13

Given: △AKM, KD ⊥ AM , AK = 6, KM = 10, m∠AKM = 93º Find: KD

Mathematics
1 answer:
storchak [24]3 years ago
6 0

Answer:

The measure of the perpendicular KD is 8.72 unit .

Step-by-step explanation:

Given as :

The triangle AKM  , with KD perpendicular to AM

The measure of side AK = 6 unit

The measure of side KM = 10 unit

The ∠ AKM = 93°

Let The measure of side KD = x unit

Now,

∵ KD ⊥ AM , KD divide the angle  ∠ AKM  equally

So,  ∠ AKD =  ∠ \dfrac{AKM}{2}

I.e  ∠ AKD =  ∠ \{93}{2}

∴  ∠ AKD = 46.5°

Now, Again

Cos Ф =  \dfrac{AK}{KD}

I.e Cos 46.5° = \dfrac{6}{KD}

I.e 0.688 = \dfrac{6}{KD}

∴ KD = \dfrac{6}{0.688}

I.e KD = 8.72 unit

Hence The measure of the perpendicular KD is 8.72 unit  Answer

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