Question:
In triangle ABC ∠A=120°. AB=AC= 10 centimetres. What is the length of BC?
A5√3
B10√3
C10+√3
D5+√3
Answer:
B. 10√3
Explanation:
In triangle ABCABC, we are given:
∠A=120
AB=AC = 10 cm (i.e., triangle ABC is isosceles with AB = AC).
We need to find the length of BC.
Step 1: Use the Law of Cosines.
The Law of Cosines states that in any triangle:
BC²=AB²+AC²−2×AB×AC×cos(∠A)
Substitute the known values:
AB=AC=10 cm,
∠A=120⁰,
cos(120⁰)= −½
So, we have:
BC²=10²+10²−2×10×10×(−½)
BC²=100+100+100=300
Thus,
BC=√300=10√3 cm.
Conclusion:
The length of BC is 10√3cm, which is approximately 17.32cm
OR
Let P be the midpoint of BC
AP = 10 × sin60
= 10 × √3/2
= 10√3/2
= 5√3
Similarly
PC = 5√3
AC = 5√3 + 5√3
= 10√3