Question

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 1

(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

Answer 2

Hi ,

*******************************************

We know that ,

In a Rhombus ,

i ) all sides are equal and opposite sides

parallel.

ii ) diagonals bisectors each other perpendicularly

iii ) Sum of the adjecent angles are

supplementary .

***********************************************

According to the problem given ,

i ) m<BTS = 110°

m<TBE = 180° - m<BTS

= 180° - 110°

= 70°

m<TBS = ( m<TBE /2 )

= 70°/2

= 35°

ii ) I( TE ) = 24

I( TA ) = I( TE )/2 = 24/2 = 12

I( BS ) = 70

I( AS ) = I( BS )/2 = 70/2 = 35

Now ,

From Right angled ∆TAS

TS² = AS² + TA²

[ By Phythogarian theorem ]

TS² = 35² + 12²

= 1225 + 144

= 1369

TS = √1369

TS = 37

Therefore ,

m<TBS = 35° ,

I( TS ) = 37

I hope this helps you.

: )