Advertisements
Advertisements
प्रश्न
In the following figure, write BC, AC, and CD in ascending order of their lengths.
Advertisements
उत्तर

In ΔABC,
∠BAC < ∠ABC
BC < AC ....( 1 )
Now, ∠ACB = 180° - ∠ABC - ∠BAC
∠ACB = 180° - 73° - 47°
∠ACB = 60°
Now, ∠ACD = 180° - ∠ACB
∠ACD = 180° - 60° = 120°
Now, in ΔACD,
∠ADC = 180° - ∠ACD - ∠CAD
∠ADC = 180° - 120° - 31°
∠ADC = 29°
Since ∠ADC < ∠CAD, we have
AC < CD ....( 2 )
From ( 1 ) and ( 2 ), we have
BC < AC < CD.
संबंधित प्रश्न
In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Arrange the sides of ∆BOC in descending order of their lengths. BO and CO are bisectors of angles ABC and ACB respectively.

Arrange the sides of the following triangles in an ascending order:
ΔABC, ∠A = 45°, ∠B = 65°.
Name the smallest angle in each of these triangles:
In ΔABC, AB = 6.2cm, BC = 5.6cm and AC = 4.2cm
Name the smallest angle in each of these triangles:
In ΔPQR, PQ = 8.3cm, QR = 5.4cm and PR = 7.2cm
ΔABC is isosceles with AB = AC. If BC is extended to D, then prove that AD > AB.
D is a point on the side of the BC of ΔABC. Prove that the perimeter of ΔABC is greater than twice of AD.
ABCD is a quadrilateral in which the diagonals AC and BD intersect at O. Prove that AB + BC + CD + AD < 2(AC + BC).
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that : PN < RN
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PR > PS
