Q5 of 31 Page 114

Using the sine and cosine tables, and if needed a calculator, do these problems.

A triangle is made by drawing angles of 50° and 65° at the ends of a line 5 centimetres long. Calculate its area.

Let ABC be the required triangle.



Draw perpendicular from AS on BC.



Let BS = x cm


⇒ SC = (5 – x) cm


In Δ BAS,


AS = BS × tan 50°


⇒ AS = x × tan 50° cm …(i)


In Δ CAS,


AS = CS × tan 65°


⇒ AS = (5 - x) × tan 65° cm …(ii)


From eq. (i) and (ii), We get :


x × tan 50° = (5 - x) × tan 65° cm


⇒ x tan 50° = 5 tan 65° - x tan 65°


⇒ x tan 50° + x tan 65° = 5 tan 65°


⇒ x (tan 50° + tan 65°) = 5 tan 65°



(From table, tan 65° = 2.144 and tan 50° = 1.191)



In Δ BAC,


Height = x = 3.21 cm and Base = 5 cm




Area (Δ) = 8.025 cm2


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