Advertisements
Advertisements
Question
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that: SN < SR
Advertisements
Solution
In ΔRTQ,
∠RTQ + ∠TQR + ∠TRQ = 180°
90° + 60° +∠TRQ = 180°
150° + ∠TRQ = 180°
∠TRQ = 180° - 150°
∠TRQ = 30°
∠TRQ = ∠SRN = 30° ....(i)
In NSR,
∠RNS + ∠SRN = 90° ....(∵∠NSR = 90°)
∠RNS + 30° = 90° ....[from (iii)]
∠RNS = 90° - 30°
∠RNS = 60° ....(ii)
∠SRN < ∠RNS ....(from (iii) and (iv))
SN < SR.
APPEARS IN
RELATED QUESTIONS
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.
D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.
“Caste inequalities are still prevalent in India.” Examine the statement.
"Issues of caste discrimination began to be written about in many printed tracts and essays in India in the late nineteenth century." Support the statement with two suitable examples.
In ΔABC, the exterior ∠PBC > exterior ∠QCB. Prove that AB > AC.
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that : PN < RN
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PR > PS
In the given figure, T is a point on the side PR of an equilateral triangle PQR. Show that PT < QT
In ΔABC, AE is the bisector of ∠BAC. D is a point on AC such that AB = AD. Prove that BE = DE and ∠ABD > ∠C.
