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, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.
“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.
Arrange the sides of the following triangles in an ascending order:
ΔDEF, ∠D = 38°, ∠E = 58°.
Prove that the perimeter of a triangle is greater than the sum of its three medians.
In ABC, P, Q and R are points on AB, BC and AC respectively. Prove that AB + BC + AC > PQ + QR + PR.
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PQ > PS
In ΔPQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.
ΔABC in a isosceles triangle with AB = AC. D is a point on BC produced. ED intersects AB at E and AC at F. Prove that AF > AE.
