Question

In the given figure, ∠ACB = 90°, ∠BDC = 90°, CD – 4 cm, BD = 3 cm, AC – 12 cm, tan A – cot A is equal to

Answer 1

Answer:

(1) It is given that line AB is tangent to the circle at A.

∴ ∠CAB = 90º (Tangent at any point of a circle is perpendicular to the radius throught the point of contact)

Thus, the measure of ∠CAB is 90º.

(2) Distance of point C from AB = 6 cm (Radius of the circle)

(3) ∆ABC is a right triangle.

CA = 6 cm and AB = 6 cm

Using Pythagoras theorem, we have

BC2=AB2+CA2⇒BC=

√

62+62

⇒BC=6

√

2

cm

Thus, d(B, C) = 6

√

2

cm

(4) In right ∆ABC, AB = CA = 6 cm

∴ ∠ACB = ∠ABC (Equal sides have equal angles opposite to them)

Also, ∠ACB + ∠ABC = 90º (Using angle sum property of triangle)

∴ 2∠ABC = 90º

⇒ ∠ABC =

90°

2

= 45º

Thus, the measure of ∠ABC is 45º.

I hope it's your helpful........

Answer 2

ABC is a right angled triangle in which ∠ABC = 90° and ∠ACB = 60°. BC is produced to D such that ∠ADB = 45°. If CD = 30 cm, what are the lengths of AB and BC?

Fully charged in 36 min - 65W SuperVOOC 2.0 in OPPO Reno4 Pro.

Let length of AB=x cm.

In right angled triangle ABD , AB/BD=tan 45°.

or. x/BD = 1 => BD=x cm.

BC= BD-CD = (x-30) cm.

In right angled triangle ABC , AB/BC= tan60°

or. x/(x-30) = √3.

or. √3.x -30√3 = x.

or. x(√3–1) = 30√3.

or. x= 30.√3/(√3–1) = 30√3×(√3+1)/(√3–1).(√3+1).

or. x =15√3(√3+1) = 15(3+√3) cm = 70.98 cm.

Thus , length of AB= x = 70.98 cm.

and length of BC= (x-30) = 40.98 cm. Answer.