∆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

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this

nothjing

step-by-step explanation: