Question

How can I solve this.
Given: △PST, m∠S=90°, M∈ segment PT,
segment PM ≅ MT,
MK ⊥ PT
m∠SPK/m∠KPM = 5/2

Find: m∠P, m∠T, m∠SKP, and m∠MKT

Answer 1

Answer:

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Answer 2

Answer:

m∠P=70°, m∠T=20°, m∠SKP=40°, and m∠MKT=70°.

Step-by-step explanation:

Given information: △PST, m∠S=90°, M∈ segment PT, segment PM ≅ MT, MK ⊥ PT, m∠SPK/m∠KPM = 5/2.

Let the measure of m∠SPK and ∠KPM are 5x° and 2x° respectively.

In triangle PKM and TKM,

$KM\cong KM$                (Common side)

$\anlge KMP\cong \angle KMT$                (MK ⊥ PT)

$PM\cong MT$           (Given)

By SAS postulate,

$\trianlge KMP\cong \triangle KMT$

$\anlge KPM\cong \triangle KTM$              (CPCTC)

$\triangle KTM=2x$

According to angle sum property, the sum of interior angles of a triangle is 180°.

Use angle sum property in triangle SPT,

$\angle P+\angle T+\angle S=180^{\circ}$

$(5x+2x)^{\circ}+(2x)^{\circ}+(90)^{\circ}=180^{\circ}$

$9x^{\circ}=180^{\circ}-90^{\circ}$

$9x^{\circ}=90^{\circ}$

$x=10$

The value of x is 10.

$\angle P=5x+2x=7x\Rightarrow 7\times 10=70^{\circ}$

$\angle T=2x\Rightarrow 2\times 10=20^{\circ}$

Therefore, m∠P=70° and m∠T=20°.

Use angle sum property in triangle SPK.

$\angle S+\angle SPK\angle SKP=180^{\circ}$

$\angle SKP=180^{\circ}-\angle S-\angle SPK$

$\angle SKP=180^{\circ}-90^{\circ}-(5x)^{\circ}$

$\angle SKP=90^{\circ}-(5\times 10)^{\circ}$

$\angle SKP=90^{\circ}-50^{\circ}=40^{\circ}$

Therefore the measure of ∠SKP is 40°.

Use angle sum property in triangle MKT.

$\angle T+\angle M+\angle K=180^{\circ}$

$20^{\circ}+90^{\circ}+\angle K=180^{\circ}$

$110^{\circ}+\angle K=180^{\circ}$

$\angle K=180^{\circ}-110^{\circ}$

$\angle K=70^{\circ}$

Therefore, the measure of ∠MKT is 70°.