Question

In Fig. 10.77, PA and PB are tangents to the circle with centre O such that ∠APB = 50°. Write the measure of ∠OAB

Answer 1

Answer:

டாபரினீர் மும்மை விதியில் நடு தனிமத்தின்அணு எடையானது முதல் மற்றும் மூன்றாம்அணு நிறையின் ------------ ஆகும்.

Step-by-step explanation:

do same as this:-

Question:-

ΔABC is an isosceles triangle in which ∠C= 90° . If AC = 6 cm, then AB=

A. 6√2 cm

B. 6 cm

C. 2√6 cm

D. 4√2cm

Answer:-

In triangle ABC

Let tge altitude be AD on BC

AC=25cm

AB=25cm

BC=14cm

Therefore DC= 1/2*BC

=1/2*14

=7

By Pythagoras theorem,

AD^2+CD^2=AC^2

AD^2+7^2=25^2

AD^2+49=625

AD^2=625-49

AD^2=576

AD=24

Therefore altitude from A on BC is 24 cm

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Answer 2

∠APB = 50°

Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3: Sum of all angles of a triangle = 180°.

By property 1,

AP = BP (tangent from P)

Therefore, ∠PAB = ∠PBA

Now,

By property 3 in ∆PAB,

∠PAB + ∠PBA + ∠APB = 180°

⇒ ∠PAB + ∠PBA = 180° – ∠APB

⇒ ∠PAB + ∠PBA = 180° – 50°

⇒ ∠PAB + ∠PBA = 130°

By property 2,

∠PAO = 90°

Now,

∠PAO = ∠PAB + ∠OAB

⇒ ∠OAB = ∠PAO – ∠PAB

⇒ ∠OAB = 90° – 65° = 25°

Hence, ∠OAB = 25°