In the given triangle ABC- - A -50- - c -50- AB -10 cm- then find the length of BC- -sin 60- -0-86- sin 50-0-76-

In the given triangle applying sine rule we have BC/sin 60^∘=AB/sin 50^∘ BC/0.86=10/0.76 BC = 8.6 × 10/0.76 BC = 11.31 cm ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Sine Rule

In the given ...

Question

In the given triangle $A B C, ∠ A = 50^{\circ}, ∠ c = 50^{\circ}$ , AB =10 cm, then find the length of BC. $[s i n 60^{\circ} = 0.86, s i n 50^{\circ} = 0.76]$

Open in App

Solution

In the given triangle applying sine rule we have

$\frac{B C}{s i n 60^{\circ}} = \frac{A B}{s i n 50^{\circ}}$

$\frac{B C}{0.86} = \frac{10}{0.76}$

BC = $\frac{8.6 \times 10}{0.76}$

BC = 11.31 cm

Suggest Corrections

Similar questions

In the given triangle $A B C, ∠ A = 50^{\circ}, ∠ c = 50^{\circ}$ , AB =10 cm, then BC is equal to $[s i n 60^{\circ} = 0.86, s i n 50^{\circ} = 0.76]$

In Triangle $A B C, A B = 15 c m, A C = 8 c m, ∠ A = 50^{\circ},$ as shown below. Find the length of BC. $[s i n 50^{\circ} = 0.76, c o s 50^{\circ} = 0.64]$

In the given figure triangle ABC is circumscribed by a circle, AB =5 cm, $∠ A = 30^{\circ}, ∠ C = 60^{\circ}$ . Find the length of side BC. $[s i n 60^{\circ} = 0.86, s i n 30^{\circ} = 0.5]$

Q. In triangle

A B C, A B = A C = 10 c m . ∠ A B C = 50^{\circ}

.
(a) Find the length of

B C

.
(b) Find the diameter of the circle.

[sin 50^{\circ} = 0.77, cos 50^{\circ} = 0.64, tan 50^{\circ} = 1.19]

In triangle ABC, $∠ C A B = 40^{\circ}$ , AC = 6 cm, AB = 7 cm. Find the length BC. $[s i n 40^{\circ} = 0.67, c o s 40^{\circ} = 0.76]$

Join BYJU'S Learning Program

In the given triangle ABC,∠A=50∘,∠c=50∘, AB =10 cm, then find the length of BC. [sin 60∘=0.86,sin 50∘=0.76]

In the given triangle applying sine rule we have BCsin 60∘=ABsin 50∘ BC0.86=100.76 BC = 8.6×100.76 BC = 11.31 cm

In the given triangle $A B C, ∠ A = 50^{\circ}, ∠ c = 50^{\circ}$ , AB =10 cm, then find the length of BC. $[s i n 60^{\circ} = 0.86, s i n 50^{\circ} = 0.76]$

In the given triangle applying sine rule we have

$\frac{B C}{s i n 60^{\circ}} = \frac{A B}{s i n 50^{\circ}}$

$\frac{B C}{0.86} = \frac{10}{0.76}$

BC = $\frac{8.6 \times 10}{0.76}$

BC = 11.31 cm