∆ABC is an equilateral triangle. Point P is on base BC such that PC = 1/3 BC. If AB = 12 cm, find AP​

Step-by-step explanation:

It is given that ABC is an equilateral triangle, then AB=BC=AC=6cm and ∠A=∠B=∠C=60°, then according to question, PC=\frac{1}{3}BC

therefore PC=2 cm.

Now, using the cosine formula in ΔAPC, we have

cos∠C= \frac{AC^{2}+PC^{2}-AP^{2}}{2(AC)(PC)}

cos60°=\frac{6^{2}+2^{2}-AP^{2}}{2(6)(2)}

\frac{1}{2}=\frac{40-AP^{2}}{24}

AP^{2}=40-12

AP^{2}=28

AP=2\sqrt{7}cm

step-by-step explanation:

consider an orange peel: if you want to try and lay it flat, you have to stretch it, we know exactly how things are being stretched or squashed at any given point. we have many different map projections because each has different patterns of some projections can even preserve certain features of the earth without

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