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2 answers:
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Answer:
<u>The measure of the arc CD = 64°</u>
Step-by-step explanation:
The rest of the question is the attached figure.
It is required to find the measure of the arc CD in degrees.
as shown at the graph
BE and AD are are diameters of circle P
And ∠APE is a right angle ⇒ ∠APE = 90°
∠APE and ∠BPE are supplementary angles
So, ∠APE + ∠BPE = 180°
∠BPE + 90 = 180°
∴ ∠BPE = 180 -90 = 90° ⇒(1)
But it is given: ∠BPE = (33k-9)° ⇒(2)
From (1) and (2)
∴ 33k - 9 = 90
∴ 33k = 90 + 9 = 99
∴ k = 99/33 = 3
The measure of the arc CD = ∠CPD = 20k + 4
By substitution with k
<u>∴ The measure of the arc CD = 20*3 + 4 = 60 + 4 = 64°</u>
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
