Question

In the figure triangle ABC angle A=120°, angle B= 30°.AC=15cm
a. What is the length of AB?
b.What is the length of BC?
Please answer the question.... No irrelevant answers needed.​

Answer 1

Answer:

Answer:

Here, I am referring to the diagram you have posted

A = 120° (given)

B = 30° (given)

we know, A + B + C = 180° [Angle Sum Property of A Triangle]

⇒ 120° + 30° + C = 180° [Replacing the given values of A and B]

⇒ 150° + C = 180°

⇒C = 30°

∴ C = B [Each 30°]

∴ ΔABC is an isosceles triangle, where AB = AC

AC = 15cm (given)

∴ AB = 15cm

From Here, I'm referring to my diagram (Its your diagram only which I have manipulated a little bit)

Now to find BC:

Drop a perpendicular from A to BC (CONSTRUCTION)

Lets call it AD

Now in ΔADC, ∠ADC = 90° (BY CONSTRUCTION)

AC = 15cm (given)

∠DAC = ∠CAB/2 = 60°

We need to find DC or CD

Sin∠DAC = CD/AC

⇒Sin60 = CD/15

⇒$\sqrt{3}$/2 = CD/15

⇒15$\sqrt{3}$/2 = CD

Similarly, BD will also be 15$\sqrt{3}$/2 [because in ΔADB, ∠DAB = 60, ∠ABD = 30, which is similar to what we solved for ΔADC]

You can calculate and try.

⇒15$\sqrt{3}$/2 = BD

Now, BC = CD + BD

BC = 15$\sqrt{3}$/2 + 15$\sqrt{3}$/2

BC = 2(15$\sqrt{3}$)/2

BC = 15$\sqrt{3}$ cm

So, we found out: AB = 15cm and BC = 15$\sqrt{3}$ cm

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