Find the measures of the following to the nearest tenth of a degree. ¯HT ∠T Solution: Given Data: WH=15cm WT=8cm ¯HT=? ∠T=? Sol: A As we know here in the following figure (HWT) is a right angle triangle, where WT is base & WH is perpendicular and HT is hypotenuses. So, according to Pythagoras theorem, (Hypotenuses)^2=(Base)^2+(Perpendicular)^2 (HT)^2=(WT)^2+(WH)^2 As here Base and perpendicular are known and to find Hypotenuses, Let, WT=a, WH=band HT=c. Hence, a^2+b^2=c^2 a=8 b=15 c=? To find c (HT) put values of a and b in the equation, a^2+b^2=c^2 8^2+15^2=c^2 c^2=8^2+15^2 c^2=64+225 c^2=289 To find ¯HT take the square root of (∴c=HT) (HT)^2=√289 ¯HT=√289 ¯HT=17Ans Sol: B As we know tan θ=Perpendicular/Base and∠T=〖tan〗^(-1)〖Perpendicular/Base〗. So, to find ∠Tput given values in the following equation ∠T=〖tan〗^(-1)〖WH/WT〗 ∠T=〖tan〗^(-1)〖15/8〗 ∠T=〖tan〗^(-1)(1.875) ∠T=61.927^∘≈61.93^∘ Ans What is the area of the given triangle? Solution: Given Data: BC
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