In Triangle Pqr Pr Is Greater Than Pq And Ps Bisects Angle Qpr P

In Triangle Pqr Pr Is Greater Than Pq And Ps Bisects In the figure, p r>p q and p s bisects ∠qp r. then ∠p sr> ∠p sq. view solution. question 5. in the given figure, pr > pq and ps bisects ∠ qpr. prove that ∠ psr > ∠ psq. . view solution. click here:point up 2:to get an answer to your question :writing hand:in adjacent figure pr pq and ps bisects angle qpr prove that angle psr. In fig 7.51, pr > pq and ps bisects ∠qpr. prove that ∠psr > ∠psq. solution: given: pr > pq and ps bisects ∠qpr. to prove: ∠psr > ∠psq. as pr > pq, ∠pqs > ∠prs (angle opposite to larger side is larger ) (1) ps is the angle bisector of ∠qpr. ∠qps = ∠rps (2) [since,ps bisects ∠qpr] ∠psr is the exterior angle of Δpqs.

Pr Pq And Ps Bisects Angle Qpr Prove That Angle Psr Psq Cl As pr > pq, ∴ ∠pqr > ∠prq (angle opposite to larger side is larger) (1) ps is the bisector of ∠qpr. ∴∠qps = ∠rps (2) ∠psr is the exterior angle of Δpqs. ∴ ∠psr = ∠pqr ∠qps (3) ∠psq is the exterior angle of Δprs. ∴ ∠psq = ∠prq ∠rps (4) adding equations (1) and (2), we obtain. Ex. 7.4, q no 5,pr greater than pq and ps bisects angle qpr. prove that angle psr greater than psq, class 9, triangles. Ex7.4, 5 in the given figure, pr > pq and ps bisects ∠qpr. prove that ∠psr >∠psq. given pr > pq, ∴ ∠pqr > ∠prq ps is the bisector of ∠qpr. ∴ ∠qps = ∠rps let ∠qps = ∠rps = x in Δ pqs, ∠psr is the exterior angle ∠psr = ∠pqr x now, ∠ pqr > ∠ prq adding x both sides ∠pqr x > ∠prq x ∠psr > ∠psq hence. Given: in $\delta pqr$, ps is a bisector of $\angle qpr$ and side pr is larger than side pq. ps is a bisector of $\angle qpr$ so it divides the angle in two equal parts. $\angle qps = \angle rps = x$ from$\delta pqs$, $\angle psr$ is the exterior angle of the triangle and we know that the exterior angle of a triangle is equal to the sum of.

Question 5 In Figure Pr Pq And Ps Bisects Qpr Prove Angle Psr Ex7.4, 5 in the given figure, pr > pq and ps bisects ∠qpr. prove that ∠psr >∠psq. given pr > pq, ∴ ∠pqr > ∠prq ps is the bisector of ∠qpr. ∴ ∠qps = ∠rps let ∠qps = ∠rps = x in Δ pqs, ∠psr is the exterior angle ∠psr = ∠pqr x now, ∠ pqr > ∠ prq adding x both sides ∠pqr x > ∠prq x ∠psr > ∠psq hence. Given: in $\delta pqr$, ps is a bisector of $\angle qpr$ and side pr is larger than side pq. ps is a bisector of $\angle qpr$ so it divides the angle in two equal parts. $\angle qps = \angle rps = x$ from$\delta pqs$, $\angle psr$ is the exterior angle of the triangle and we know that the exterior angle of a triangle is equal to the sum of. We now compare the two triangles $\delta pqs$ and $\delta psr$ . one of the angles of the two triangles being the same and one angle being greater than the other $\left( \angle pqr>\angle prq \right)$ , we must then have the remaining angle smaller than the other of the second triangle. complete step by step answer:. Given: $pr > pq$ and $ps$ bisects $\angle qpr$. to do: we have to prove that $\angle psr > \angle psq$. solution: let us consider $\triangle pqr$ we have,.