WORK OUT THE RELATIVE FREQUENCY AS A DECIMAL OF THROWING A 4
1 answer:
You might be interested in
Answer:
∠MSR is right angle
Step-by-step explanation:
It is given that RHOM is a Rhombus
Also ΔHOM, ΔMHR, ΔRHO and ΔOMR are isosceles triangles
Let us take ΔMSR and ΔRSH
∠MRS =∠HRS ( since it is given that the diagonal RO bisects ∠R)
∠RMS =∠RHS ( since Δ MRH is isosceles triangle )
RS = RS ( common side )
By AAS congruency rule ΔMSR ≅ ΔHSR
so we have
∠MSR=∠RSH ( corresponding parts of congruent triangles are congruent)
also we have
∠MSR +∠RSH =180° ( supplementary angles)
∠MSR +∠MSR=180° ( since ∠MSR=∠RSH)
2∠MSR= 180°
∠MSR =90°
Hence ∠MSR is right angle
