Ex 8.1, 10
In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Given
PR + QR = 25 cm
Let QR = x
Thus,
PR + QR = 25 cm
PR = 25 – QR
PR = 25 – x
In right triangle PQR,
Using Pythagoras theorem
(Hypotenuse)2 = (Height)2 + (Base)2
PR2 = PQ2 + QR2
(25 – x)2 = 52 + x2
(25)2 + 𝑥2 − 2 × 25 × 𝑥=25+𝑥2
625 + x2 – 50x = 25 + x2
625 + x2 – 50x – 25 – x2 = 0
x2 – x2 – 50x + 625 – 25 = 0
− 50𝑥 + 600 = 0
− 50𝑥 = −600
x = (− 600)/(− 50)
x = 12
Hence,
QR = x = 12 cm
PR = 25 – x
= 25 – 12
= 13 cm
sin P = (𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑃 )/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
= 𝑄𝑅/𝑃𝑅
= 𝟏𝟐/𝟏𝟑
cos P = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑃 )/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
= 𝑃𝑄/𝑃𝑅
= 𝟓/𝟏𝟑
And
tan P = (𝒔𝒊𝒅𝒆 𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒕𝒐 𝑷 )/(𝒔𝒊𝒅𝒆 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒕𝒐 𝑷)
tan P = 𝑄𝑅/𝑃𝑄
tan P = 12/5
tan P = 𝟏𝟐/𝟓
Alternatively,
tan P = 𝒔𝒊𝒏𝐏/𝒄𝒐𝒔𝐏
tan P = (12/13)/(5/13)
tan P = 𝟏𝟐/𝟓

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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