Advertisements
Advertisements
प्रश्न
Arrange the sides of ∆BOC in descending order of their lengths. BO and CO are bisectors of angles ABC and ACB respectively.
Advertisements
उत्तर
∠BAC = 180° - ∠BAD = 180° - 137° = 43°
∠ABC = 180° - ∠ABE = 180° - 106° = 74°
Thus, in ΔABC,
∠ACB = 180° - ∠BAC - ∠ABC
⇒ ∠ACB = 180° - 43° - 74° = 63°
Now, ∠ABC = ∠OBC + ∠ABO
⇒ ∠ABC = 2∠OBC ....( OB is biosector of ∠ABC )
⇒ 74° = 2∠OBC
⇒ ∠OBC = 37°
Similarly,
∠ACB = ∠OCB + ∠ACO
⇒ ∠ACB = 2∠OCB ...( OC is bisector of ACB )
⇒ 63° = 2∠OCB
⇒ ∠OCB = 31.5°
Now, in ΔBOC,
∠BOC = 180° - ∠OBC - ∠OCB
⇒ ∠BOC = 180° - 37° - 31.5°
⇒ ∠BOC = 111.5°
Since, ∠BOC> ∠OBC > ∠OCB, we have
BC > OC > OB
APPEARS IN
संबंधित प्रश्न
In a huge park people are concentrated at three points (see the given figure):
A: where there are different slides and swings for children,
B: near which a man-made lake is situated,
C: which is near to a large parking and exit.
Where should an ice-cream parlour be set up so that maximum number of persons can approach it?
(Hint: The parlor should be equidistant from A, B and C)
In the following figure, ∠BAC = 60o and ∠ABC = 65o.
Prove that:
(i) CF > AF
(ii) DC > DF
In the following figure ; AC = CD; ∠BAD = 110o and ∠ACB = 74o.
Prove that: BC > CD.
Name the greatest and the smallest sides in the following triangles:
ΔABC, ∠ = 56°, ∠B = 64° and ∠C = 60°.
Arrange the sides of the following triangles in an ascending order:
ΔDEF, ∠D = 38°, ∠E = 58°.
Name the smallest angle in each of these triangles:
In ΔPQR, PQ = 8.3cm, QR = 5.4cm and PR = 7.2cm
Prove that the perimeter of a triangle is greater than the sum of its three medians.
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that : PN < RN
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PQ > PS
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PR > PS
