AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

In triangle $ABC, AB = AC = 10 cm . ∠ ABC = 50^{\circ}$ .
(a). Find the length of BC
(b). Find the diameter of the circle.

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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

13.14 cm

12.14 cm

15.14 cm

17.14 cm

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The figure for the given problem can be given as,
Question Image

Here,

AB = AC = 10 cm,

△ ABC

is an isosceles triangle and hence the base angles are equal.
Now, the angles,

m ∠ B = m ∠ C = 50^{\circ}

From angle sum property,

m ∠ A + m ∠ B + m ∠ C = 180^{\circ}

m ∠ A + 50^{\circ} + 50^{\circ} = 180^{\circ}

m ∠ A + 100^{\circ} = 180^{\circ}

m ∠ A = 80^{\circ}

Now, perpendicular which is drawn from vertex of the isosceles triangle to base bisects this base in two equal parts.
(a) In this part right triangle

△ ABD,

\cos 50^{\circ} = \frac{BD}{AB}

0.64 = \frac{BD}{10}

\Rightarrow 0.64 \times 10 = BD

\Rightarrow BD = 6.4 cm

So,

BC = BD + DC = 25

BD = 2 \times 6.4

BD = 12.8 cm

(b) For this part we have,

OA = OB = OC

= circumradius of

△ ABC

.
Suppose,
A = area of

△ ABC

R = circumradius.
Here,

A = \frac{abc}{4 R}

, (a, b, c =sides of the triangle)
In the

△ ABC,

we have,

\tan 50 = \frac{AD}{BD}

\Rightarrow 1.19 = \frac{AD}{6.4}

\Rightarrow AD = 1.19 \times 6.4

= 7.616 cm

Area

(△ ABC) = \frac{1}{2} \times base \times height

= \frac{1}{2} \times BC \times AD

= \frac{1}{2} \times 12.8 \times 7.616

= 48.7424 sq . cm

As,

A = \frac{abc}{4 R}

, thus, we get,

R = \frac{abc}{4 A}

R = \frac{10 \times 10 \times 12.8}{4 \times 48.7424}

R = 6.57 cm

So, the value of diameter of circle

D = 2 R

D = 2 \times 6.57

D = 13.14 cm

So, option 1 is correct.

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!