1.

Diagonals of a rhombus BEST intersect at A.(i) If m∠BTS = 110°, then find m∠TBS(ii) If l(TE) = 24, l(BS) = 70, then find l(TS)= ?

Answer»
(i) now, ∆TBS ,
BT = TS
so, ∠TBS = ∠TSB ------(1)
Given, ∠BTS = 110°
we KNOW, sum of all angles of triangle is 180°
so, ∠BTS + ∠TBS + ∠TSB = 180°
=> 110° + 2(∠TBS) = 180° [ from equation (1)]
=>2(∠TBS) = 180° - 110° = 70°
=> ∠TBS = 35°
hence, ∠TBS = 35°

(ii) Given, length of TE = 24
length of BS = 70
we know, diagonals of rhombus are perpendicular bisector to each other.
so, BA = AS = BS/2 = 35
similarly, TA = AE = TE/2 = 12
now, ∆ATS is right - angled triangle.
from Pythagorean theorem,
TS² = TA² + AS²
= 35² + 12² = 1225 + 144 = 1369
TS = \sqrt{1369}
TS = 37


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